A spinner has 10 equally sized sections, 8 of which are gray and 2 of which are blue. The spinner is spun twice. What is the probability that the first spin lands on gray and the second spin lands on blue ? Write your answer as a fraction in simplest form.

Answers

Answer 1

Answer:

4/25

Step-by-step explanation:

The probability the first spin lands on gray is 8/10 = 4/5.

The probability the second spin lands on blue is 2/10 = 1/5.

The probability of both events is 4/5 × 1/5 = 4/25.


Related Questions

normally distributed with an unknown population mean and a population standard deviation of 4.5 points. A random sample of 45 scores is taken and gives a sample mean of 84. Find a 90% confidence interval

Answers

Answer:

= ( 82.90, 85.10) points

Therefore at 90% confidence interval (a,b)= ( 82.90, 85.10) points

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 84

Standard deviation r = 4.5

Number of samples n = 45

Confidence interval = 90%

z(at 90% confidence) = 1.645

Substituting the values we have;

84+/-1.645(4.5/√45)

84+/-1.645(0.670820393249)

84+/-1.10

= ( 82.90, 85.10) points

Therefore at 90% confidence interval (a,b)= ( 82.90, 85.10) points

divide the following polynomials ( 9 x 4 + 3 x 3 y − 5 x 2 y 2 + x y 3 ) ÷ ( 3 x 2 + 2 x 2 y − x y 2 )

Answers

Answer:

2(-2y+9)/3+y

Step-by-step explanation:

I need help with this one

Answers

Answer:

2 2/3

Step-by-step explanation:

which is pattern 12,24,36,48

Answers

t(n)=n×12, where n=the nth term in the sequence and where 12=a constant. always by 12

Answer:

multiples of 12

Step-by-step explanation: when looking at the GCF, the answer is 12

What’s the correct answer for this question?

Answers

Answer:

C

Step-by-step explanation:

A cylinder is formed when rotating the 3-D figure around y-axis

Suppose cattle in a large herd have a mean weight of 3181lbs and a standard deviation of 119lbs. What is the probability that the mean weight of the sample of cows would differ from the population mean by greater than 11lbs if 49 cows are sampled at random from the herd

Answers

Answer:

51.56% probability that the mean weight of the sample of cows would differ from the population mean by greater than 11lbs if 49 cows are sampled at random from the herd

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

[tex]\mu = 3181, \sigma = 119, n = 49, s = \frac{119}{\sqrt{49}} = 17[/tex]

What is the probability that the mean weight of the sample of cows would differ from the population mean by greater than 11lbs if 49 cows are sampled at random from the herd

Lower than 3181 - 11 = 3170 lbs or greater than 3181 + 11 = 3192 lbs. Since the normal distribution is symmetric, these probabilities are equal. So i will find one of them, and multiply by 2.

Probability of mean weight lower than 3170 lbs:

This is 1 subtracted by the pvalue of Z when X = 3170. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{3170 - 3181}{17}[/tex]

[tex]Z = -0.65[/tex]

[tex]Z = -0.65[/tex] has a pvalue of 0.2578

2*0.2578 = 0.5156

51.56% probability that the mean weight of the sample of cows would differ from the population mean by greater than 11lbs if 49 cows are sampled at random from the herd

help asap, will get branliest !!​

Answers

Answer:

D

Step-by-step explanation:

Lines EF and GH are already parallel. Translating them 2 units to the side without changing how far apart they are vertically means they won't intersect and will remain the same distance apart.

Answer:

D

Step-by-step explanation:

They are parallel lines

Graph the line with slope -1/3 and y -intercept 6 .

Answers

Answer:

plot a point at 6 up from (0,0) and then go down one and over three places then plot another point- and so on - and so on

Step-by-step explanation:

To graph the line using the slope and intercept, first understand what the slope and intercept mean:

Slope is how steep or flat the line appears on the graph.

A very high or low slope (100 or -100) will be very steep on the graph.A slope very close to zero (0.0001 or -0.0001) will be very flat on the graph.A positive slope will travel northeast and southwest (for linear equations).A negative slope will travel northwest and southeast (for linear equations).

The y-intercept is the point at which the line hits the y-axis. In this equation, the line hits the y-axis at positive 6, which means that the point is (0, 6).

You can use a method called "rise over run" to graph. The slope is negative one over three, so the line will "rise" negative one units after "running" three units.

So, for every one unit down, the line will travel three units to the right.

Graph this from the point (0, 6), your y-intercept, and plot the points according to the slope:

Translate to a system of equations: Twice a number plus three times a second number is negative one. The first number plus four times the second number is two.

Answers

Answer:

work is shown and pictured

A city has just added 100 new female recruits to its police force. The city will provide a pension to each new hire who remains with the force until retirement. In addition, if the new hire is married at the time of her retirement, a second pension will be provided for her husband. A consulting actuary makes the following assumptions: (i) Each new recruit has a 0.4 probability of remaining with the police force until retirement. (ii) Given that a new recruit reaches retirement with the police force, the probability that she is not married at the time of retirement is 0.25. (iii) The events of different new hires reaching retirement and the events of different new hires being married at retirement are all mutually independent events. Calculate the probability that the city will provide at most 90 pensions to the 100 new hires and their husbands. (A) 0.60 (B) 0.67 (C) 0.75 (D) 0.93 (E) 0.99

Answers

Answer:

E) 0.99

Step-by-step explanation:

100 recruits x 0.4 chance of retiring as police officer = 40 officers

probability of being married at time of retirement = (1 - 0.25) x 40 = 30 officers

each new recruit will result in either 0, 1 or 2 new pensions

0 pensions when the recruit leaves the police force (0.6 prob.)1 pension when the recruit stays until retirement but doesn't marry (0.1 prob.)2 pensions when the recruit stays until retirement and marries (0.3 prob.)

mean = µ = E(Xi) = (0 x 0.6) + (1 x 0.1) + (2 x 0.3) = 0.7

σ²  = (0² x 0.6) + (1² x 0.1) + (2² x 0.3) - µ² = 0 + 0.1 + 1.2 - 0.49 = 0.81

in order for the total number of pensions (X) that the city has to provide:

the normal distribution of the pension funds = 100 new recruits x 0.7 = 70 pension funds

the standard deviation = σ = √100 x √σ² = √100 x √0.81 = 10 x 0.9 = 9

P(X ≤ 90) = P [(X - 70)/9] ≤ [(90 - 70)/9] =  P [(X - 70)/9] ≤ 2.22

z value for 2.22 = 0.9868 ≈ 0.99

In a completely randomized design involving three treatments, the following information is provided: Treatment 1 Treatment 2 Treatment 3 Sample Size 5 10 5 Sample Mean 4 8 9 The overall mean for all the treatments is a. 7.00 b. 6.67 c. 7.25 d. 4.89

Answers

Answer:

c. 7.25

Step-by-step explanation:

Given the following information from an experiment:

[tex]\left\begin{array}{ccc}&$Sample Size&$Sample Mean \\$Treatment 1&5&4\\$Treatment 2&10&8\\$Treatment 3&5&9\end{array}\right[/tex]

Total Sample Size =5+10+5=20

Therefore, the overall mean

[tex]=\dfrac{(5 \times 4)+ (10 \times 8) + (5 \times 9)}{20} \\=\dfrac{145}{20}\\\\=7.25[/tex]

what is the common ratio of the geometric sequence below ?
-96,48,-24,12,-6... ​

Answers

Answer:

r = - [tex]\frac{1}{2}[/tex]

Step-by-step explanation:

The common ratio r is the ratio between consecutive terms in the sequence.

r = [tex]\frac{48}{-96}[/tex] = [tex]\frac{-24}{48}[/tex] = [tex]\frac{12}{-24}[/tex] = [tex]\frac{-6}{12}[/tex] = - [tex]\frac{1}{2}[/tex]

Answer:

-1/2 or b on edge

Step-by-step explanation:

Let two cards be dealt successively, without replacement, from a standard 52-card deck. Find the probability of the event. The first card is a queen and the second is a seven

Answers

Answer: 4 / 663

Step-by-step explanation:

There are 4 queens in a deck of 52 cards.

Probability = 4/52 = 1/13

There are 4 sevens

Probability = 4/51

Total probability = 1/13 x 4/51 = 4 / 663

The probability of drawing a queen first and a seven-second is 3/613.

What is probability?

Probability is defined as the possibility of an event being equal to the ratio of the number of favorable outcomes and the total number of outcomes.

There are 4 queens in a standard deck, and once one queen is drawn, there are 51 cards left, including 3 sevens.

So, the probability of drawing a queen first is 4/52 or 1/13, and the probability of drawing a seven-second is 3/51.

By the multiplication rule of probability, multiply the probabilities of each event occurring:

P(Queen and Seven) = P(Queen) × P(Seven after Queen)

P(Queen and Seven) = (1/13) × (3/51)

P(Queen and Seven) = 3/613

Thus, the probability of drawing a queen first and a seven-second is 3/613.

Learn more about the probability here:

brainly.com/question/11234923

#SPJ2

The following data represent the number of flash drives sold per day at a localcomputer shop and their prices.Price Units Sold34 336 432 635 530 938 240 1a. Develop the estimated regression equation that could be used to predict thequantity sold given the price. Interpret the slope.b. Did the estimated regression equation provide a good fit? Explain.c. Compute the sample correlation coefficient between the price and the number offlash drives sold. Use a= 0.01 to test the relationship between price and units sold.d. How many units can be sold per day if the price of flash drive is set to $28.

Answers

Answer:

a)3145 x 0.01 = 31.45  3145- 31.45 = 3113.55

Compute the sample correlation 3113.55 -? we find the least square pressing at least 15x on the calculator then minus this from 3113.55 to find a better fit and minimum regression.

We add the differences of units then divide by distribution as seen below.

b) unsure.

c) = (see below) just test each number shown unit sold per day / price then x can show the differences in each number from day 1 to day 2.

d) = 16 sold.

Step-by-step explanation:

a) We count the units up and deduct from it from the equation p is recognized as units sold. R1 is cost R2 is total days.

b) The line of best fit is described by the equation ŷ = bX + a, where b is the slope of the line and a is the intercept (i.e., the value of Y when X = 0).

c) r 2= decimal ; the regression equation has accounted for percentage of the total sum of squares. You cna do this one.

d) = 16 sold at $28 each. - Why ? We using 7 day data and prove a how many units can be sold p/d if the price of flash drive is set to $28 each per unit.

Day 1 = 34  /  28  =  1      =    1.21428571429 = 1 no difference day prior.

Day 2 = 336   / 28   = 12  = 12 = difference day prior is 11

Day 3 = 432  / 28  = 15    =  15.4285714286   = 15 difference day prior is 3

Day 4 = 635  / 28 =  23   =  22.6785714286  = 23 difference day prior is 8

Day 5  = 530  /  28 = 19    =  18.9285714286  = 19 difference day prior is minus - 4

Day 6 = 938  / 28  =  34   = 33.5 = 34 difference day prior is 15

Day 7 = 240  / 28  =   9    =   8.57142857143 = 9 difference day prior is minus -25

Total days 7 = Total revenue / price = average units sold

Average units sold total = 1+ 12+15 +23 +19+34+9 = 113 rounded.

Average units sold total = 1.21428571429 + 12 + 15.4285714286

+ 22.6785714286

+18.9285714286

+ 33.5

+ 8.57142857143 =  112.321428572 units sold weekly when priced at $28

To answer D we divide this by 7  to show;

112.321428572/ 7 = 16.0459183674

Daily units sold = 16

m is directly proportional to r squared when r=2 m=14 work out the value of r when m = 224

Answers

Answer:

32

Step-by-step explanation:

r:m

2:14

1:7

m=224

r=224 divided by 7

224/7=32

Edit: unless it is proportional to r^2 in which case it is a different answer

Answer:

m=504

Step-by-step explanation:

Any help would be great

Answers

Answer:

30%

Step-by-step explanation:

fat ÷ total

15 ÷ 50

.3

30%

Answer:

30%

Step-by-step explanation:

To find the percent from fat, take the calories from fat and divide by the total

15/50

.3

Multiply by 100%

30%

How can you use mathematics to help scientists explore Martian Craters ? 

Answers

Answer:

Mathematics could make scientists to have a preliminary understanding of the dimensions, perimeters, areas and volumes of different craters on Mars.

Step-by-step explanation:

Martian Craters are series of craters formed on the surface of Mars. The study of a planets crater gives an understanding of the properties of matter that lies under the crater.

Mathematics can be applied to determine the dimensions, perimeter, area and volume of the features of a crater using appropriate conversions and theorems.

The Pi in the sky theorem can be applied to determine the area and perimeter, even volume of different craters on the Mars surface. Also, eingenfunction expansion theorem gives a preliminary knowledge of the craters.

By measurements and conversions processes, the features of Martian crater could be studied from images.

Consider the following set of sample data.
18 26 30 42 50 52 52 76 78 84
For the given data, the mean is_______, the median is________, and the mode is_______.
Suppose the value 76 in the data is mistakenly recorded as 55 instead of 76. For the sample with this error, the mean is_________, the median is______, and the mode is_______. The mean_____, the median_______, and the mode______. Suppose the value 76 in the original sample is inadvertently removed from the sample. For the sample with this value removed, the mean is_______, the median is_______, and the mode is________. The mean_________, the median_______, and the mode________.

Answers

Answer:

For the given data, the mean is 50.8, the median is 51, and the mode is 52.

For the sample with this error, the mean is 48.7, the median is 51, and the mode is 52.

For the sample with this value removed, the mean is 43.2, the median is 50, and the mode is 52.

Step-by-step explanation:

We are given the following set of sample data below;

18, 26, 30, 42, 50, 52, 52, 76, 78, 84.

The formula for calculating mean is given by;

         Mean  =  [tex]\frac{\text{Sum of all data values}}{\text{Total number of observations}}[/tex]

                     =  [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 76+ 78+ 84}{10}[/tex]  

                     =  [tex]\frac{508}{10}[/tex]  =  50.8

For calculating median, we have to observe that the number of observations (n) in our data is even or odd, i.e;

If n is odd, then the formula for calculating median is given by;

                    Median  =  [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]

If n is even, then the formula for calculating median is given by;

                    Median  =  [tex]\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}[/tex]

Now, here in our data the number of observations is even, i.e. n = 10.

So, Median  =  [tex]\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{(\frac{10}{2})^{th}\text{ obs.} +(\frac{10}{2}+1)^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{5^{th}\text{ obs.} +6^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{50+52 }{2}[/tex]  =  51

A Mode is a value that appears the maximum number of times in our data.

In our data, the value 52 is appera]ing maximum number of times, i.e. 2 times which means that mode of our data is 52.

Now, suppose the value 76 in the data is mistakenly recorded as 55 instead of 76. For the sample with this error,

Mean will be changed as value has been changed.

            New Mean  =   [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 55+ 78+ 84}{10}[/tex]  

                                =  [tex]\frac{487}{10}[/tex]  =  48.7

There will be no change in median because there is no change in the 5th and 6th observation of the data.

Also, there will be no change in mode as stiil 52 appears maximum number of times in our data.

Now, suppose the value 76 in the original sample is inadvertently removed from the sample. For the sample with this value removed,

Mean will be changed as value has been removed from data.

            New Mean  =   [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 78+ 84}{9}[/tex]  

                                =  [tex]\frac{432}{10}[/tex]  =  43.2

Median will also get changed because the number of observation is now odd, i.e. n = 9

            So, Median  =  [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]

                                 =  [tex](\frac{9+1}{2})^{th} \text{ obs.}[/tex]

                                 =  [tex]5^{th} \text{ obs.}[/tex] = 50

Also, there will be no change in mode as stiil 52 appears maximum number of times in our data.

Line segment ON is perpendicular to line segment ML
What is the length of chord ML?
0
20 units
24 units
26 units
30 units
13
P
8
M
N
Mark this and return

Answers

Answer:

The correct answer is B (24 units)

Step-by-step explanation:

What is the center of the circle?

Answers

Answer:The point from which circle is drawn is called center of circle.

Step-by-step explanation:I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thnk me...

HELP PLEASE!!
NEED ANSWER ASAP!!!

A farmer in China discovers a mammal
hide that contains 54% of its original

Find age of the mammal hide to the nearest year.

amount of C-14
N=N0e^-kt
N = Noe
No = inital amount of C-14 (at time t = 0)
N = amount of C-14 at time t
k = 0.0001
t = time, in years

Answers

Answer:

6163.2 years

Step-by-step explanation:

A_t=A_0e^{-kt}

Where

A_t=Amount of C 14 after “t” year

A_0= Initial Amount

t= No. of years

k=constant

In our problem we are given that A_t is 54% that is if A_0=1 , A_t=0.54

Also , k=0.0001

We have to find t=?

Let us substitute these values in the formula

0.54=1* e^{-0.0001t}

Taking log on both sides to the base 10 we get

log 0.54=log e^{-0.0001t}

-0.267606 = -0.0001t*log e

-0.267606 = -0.0001t*0.4342

t=\frac{-0.267606}{-0.0001*0.4342}

t=6163.20

t=6163.20 years

PLEASE MARK BRAINLY

Solve the problem.
If a boat uses 25 gallons of gas to go 73 miles, how many miles
can the boat travel on 75 gallons of gas?
24 mi
438 mi
219 mi
239 mi

Answers

Answer:

For this problem we can use the following proportional rule:

[tex] \frac{73 mi}{25 gal}= \frac{x}{75 gal}[/tex]

Where x represent the number of miles that we can travel with 75 gallons. For this case we can use this proportional rule since by definition [tex] D=vt[/tex]. If we solve for x we got:

[tex] x =75 gal (\frac{73 mi}{25 gal}) =219mi[/tex]

And the best answer would be:

219 mi

Step-by-step explanation:

For this problem we can use the following proportional rule:

[tex] \frac{73 mi}{25 gal}= \frac{x}{75 gal}[/tex]

Where x represent the number of miles that we can travel with 75 gallons. For this case we can use this proportional rule since by definition [tex] D=vt[/tex]. If we solve for x we got:

[tex] x =75 gal (\frac{73 mi}{25 gal}) =219mi[/tex]

And the best answer would be:

219 mi

Mist (airborne droplets or aerosols) is generated when metal-removing fluids are used in machining operations to cool and lubricate the tool and work-piece. Mist generation is a concern to OSHA, which has recently lowered substantially the workplace standard. The article "Variables Affecting Mist Generation from Metal Removal Fluids" (Lubrication Engr., 2002: 10-17) gave the accompanying data on x = fluid flow velocity for a 5% soluble oil (cm/sec) and y = the extent of mist droplets having diameters smaller than some value:
x: 89 177 189 354 362 442 965
y: .40 .60 .48 .66 .61 .69 .99
a. Make a scatterplot of the data. By R.
b. What is the point estimate of the beta coefficient? (By R.) Interpret it.
c. What is s_e? (By R) Interpret it.
d. Estimate the true average change in mist associated with a 1 cm/sec increase in velocity, and do so in a way that conveys information about precision and reliability.
e. Suppose the fluid velocity is 250 cm/sec. Find the mean of the corresponding y in a way that conveys information about precision and reliability. Use 95% confidence level. Interpret the resulting interval. By hand, as in part d.
f. Suppose the fluid velocity for a specific fluid is 250 cm/sec. Predict the y for that specific fluid in a way that conveys information about precision and reliability. Use 95% prediction level. Interpret the resulting interval. By hand, as in part d.

Answers

Answer:

Step-by-step explanation:

a) image attached

b) Lets do the analysis in R , the complete R snippet is as follows

x<- c(89,177,189,354,362,442,965)

y<- c(.4,.6,.48,.66,.61,.69,.99)

# scatterplot

plot(x,y, col="red",pch=16)

# model

fit <- lm(y~x)

summary(fit)

#equation is

#y = 0.4041 + 0.0006211*X

# beta coeffiecients are

fit$coefficients

coef(summary(fit))[, "Std. Error"]

# confidence interval of slope

confint(fit, 'x', level=0.95)

The results are

> summary(fit)

Call:

lm(formula = y ~ x)

Residuals:

1 2 3 4 5 6 7

-0.05940 0.08595 -0.04151 0.03602 -0.01895 0.01136 -0.01346

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 4.041e-01 3.459e-02 11.684 8.07e-05 ***

x 6.211e-04 7.579e-05 8.195 0.00044 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.05405 on 5 degrees of freedom

Multiple R-squared: 0.9307,   Adjusted R-squared: 0.9168 # model is able to capture 93% of the variation of the data

F-statistic: 67.15 on 1 and 5 DF, p-value: 0.0004403 , p value is less than 0.05 , hence model as a whole is significant

> fit$coefficients

(Intercept) x

0.4041237853 0.0006210758

> coef(summary(fit))[, "Std. Error"]

(Intercept) x

3.458905e-02 7.579156e-05

> confint(fit, 'x', level=0.95)

2.5 % 97.5 %

x 0.0004262474 0.0008159042

c)

> x=c(89,177,189,354,362,442,965)

> y=c(0.40,0.60,0.48,0.66,0.61,0.69,0.99)

>

> ### linear model

> model=lm(y~x)

> summary(model)

Call:

lm(formula = y ~ x)

Residuals:

1 2 3 4 5 6 7

-0.05940 0.08595 -0.04151 0.03602 -0.01895 0.01136 -0.01346

Coefficients:

Estimate Std. Error t value Pr(>|t|)    

(Intercept) 4.041e-01 3.459e-02 11.684 8.07e-05 ***

x 6.211e-04 7.579e-05 8.195 0.00044 ***

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.05405 on 5 degrees of freedom

Multiple R-squared: 0.9307, Adjusted R-squared: 0.9168

F-statistic: 67.15 on 1 and 5 DF, p-value: 0.0004403

s_e is the Residual standard error from the model and its estimated value is 0.05405. s_e is the standard deviation of the model.  

d) 95% confidence interval

> confint(model, confidence=0.95)

2.5 % 97.5 %

(Intercept) 0.3152097913 0.4930377793

x 0.0004262474 0.0008159042

Comment: The estimated confidence interval of slope of x does not include zero. Hence, x has the significant effect on y at 0.05 level of significance.

 e)

> predict(model, newdata=data.frame(x=250), interval="confidence", level=0.95)

fit lwr upr

1 0.5593927 0.5020485 0.616737

f)

> predict.lm(model, newdata=data.frame(x=250), interval="prediction", level=0.95)

fit lwr upr

1 0.5593927 0.4090954 0.7096901

A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the gulcose is added, it is converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at the time. Thus a model for the concentration C=C(t) of the glucose solution in the bloodstream is
dC/dt=r-kC
Where r an dk are positive constants.
1. Suppose that the concentration at time t=0 is C0. Determine the concentration at any time t by solving the differential equation.
2. Assuming that C0

Answers

Answer:

[tex]C(t) =\dfrac{ r}{k} - \left (\dfrac{r-kC_{0}}{k} \right )e^{ -kt}[/tex]

[tex]C(t) =\dfrac{ r}{k}- e^{ -kt}[/tex]   ,thus, the  function is said to be an increasing function.

Step-by-step explanation:

Given that:

[tex]\dfrac{dC}{dt}= r-kC[/tex]

[tex]\dfrac{dC}{r-kC}= dt[/tex]

Taking integration on both sides ;

[tex]\int\limits\dfrac{dC}{r-kC}= \int\limits \ dt[/tex]

[tex]- \dfrac{1}{k}In (r-kC)= t +D[/tex]

[tex]In(r-kC) = -kt - kD \\ \\ r- kC = e^{-kt - kD} \\ \\ r- kC = e^{-kt} e^{ - kD} \\ \\r- kC = Ae^{-kt} \\ \\ kC = r - Ae^{-kt} \\ \\ C = \dfrac{r}{k} - \dfrac{A}{k}e ^{-kt} \\ \\[/tex]

[tex]C(t) =\frac{ r}{k} - \frac{A}{k}e^{ -kt}[/tex]

here;

A is an integration constant

In order to determine A, we have [tex]C(0) = C0[/tex]

[tex]C(0) =\frac{ r}{k} - \frac{A}{k}e^{0}[/tex]

[tex]C_0 =\frac{r}{k}- \frac{A}{k}[/tex]

[tex]C_{0} =\frac{ r-A}{k}[/tex]

[tex]kC_{0} =r-A[/tex]

[tex]A =r-kC_{0}[/tex]

Thus:

[tex]C(t) =\dfrac{ r}{k} - \left (\dfrac{r-kC_{0}}{k} \right )e^{ -kt}[/tex]

2. Assuming that C0 < r/k, find lim t→[infinity] C(t) and interpret your answer

[tex]C_{0} < \lim_{t \to \infty }C(t) \\ \\C_0 < \dfrac{r}{k} \\ \\kC_0 <r[/tex]

The equation for C(t) can  be rewritten as :

[tex]C(t) =\dfrac{ r}{k} - \left (\dfrac{r-kC_{0}}{k} \right )e^{ -kt}C(t) =\dfrac{ r}{k} - \left (+ve \right )e^{ -kt} \\ \\C(t) =\dfrac{ r}{k}- e^{ -kt}[/tex]

Thus;  the  function is said to be an increasing function.

One number is 4 plus one half of another number. Their sum is 31. Find the numbers.

Answers

Answer:

18, 13

Step-by-step explanation:

x=4+1/2y

x+y=31

4+1/y+y=31

3/2y=27

y=18

x=31-18=13

Answer:

13 & 18

Step-by-step explanation:

Create the formulas:

0.5x+4=y

x+y=31

0.5x+4=y

Multiply both sides by 2

x+8=2y

x+y=31

Subtract 31 from both sides

x+y-31=0

Subtract y from both sides

x-31= -y

Multiply both sides by -1

-x+31=y

Multiply both sides by 2

-2x+62=2y

Combine equations:

-2x+62=x+8

Add 2x to both sides

62=3x+8

Subtract 8 from both sides

3x=54

Divide both sides by 3

x=18

0.5x+4=y

Subtract y from both sides

0.5x-y+4=0

Subtract 0.5x from both sides

-y+4= -0.5x

Multiply both sides by -1

y-4=0.5x

Multiply both sides by 2

2y-8=x

x+y=31

Subtract y from both sides

x= -y+31

Combine equations:

2y-8= -y+31

Add y to both sides

3y-8=31

Add 8 to both sides

3y=39

Divide both sides by 3

y=13

In 2018, the number of students at The Villages High School was 975 and is increasing at a rate of 2.5% per year. Write and use an exponential growth function to project the populating in 2025. Round to the nearest whole number. Help plzzz

Answers

Answer:

[tex]A(t)=975(1.025)^t[/tex]

In 2025,the number of students at the villages high school=1159

Step-by-step explanation:

We are given that in 2018

Number of students at the villages high school=975

Increasing rate,r=2.5%=0.025

We have to write and use of exponential growth function to project the populating in 2025.

[tex]A_0=975,t=0[/tex]

According to question

Number of students at the villages high School is given by

[tex]A(t)=A_0(1+r)^t[/tex]

Substitute the values

[tex]A(t)=975(1+0.025)^t=975(1.025)^t[/tex]

t=7

Substitute the value

Then, the number of students at the villages high school in 2025

[tex]A(7)=975(1.025)^7=1158.96\approx 1159[/tex]

Answer:

1,159 students

Step-by-step explanation:

the exponential growth rate formula:

A = P ( 1 + r)ⁿ

A = amount after growth = ??P = current/original amount = 975 studentsr = yearly growth rate = 2.5% or 0.025n = number of years = 2025 - 2018 = 7

Pop. 2025 = 975 (1 + 0.025)⁷

Pop. 2025 = 975 x 1.025⁷ = 1,158.97 ≈ 1,159 students

What is the simplified value of the exponential expression 27 1/3?
1/3
1/9
3
9

Answers

Answer:3

Step-by-step explanation:

Answer:

C.3

Step-by-step explanation:

Which expression is equivalent to log Subscript 8 Baseline 4 a (StartFraction b minus 4 Over c Superscript 4 Baseline EndFraction)?

Answers

Answer:

[tex]\log_84+\log_8a+\log_8(b-4)-4\log_8c[/tex].

Step-by-step explanation:

The given expression is

[tex]\log_84a\left(\dfrac{b-4}{c^4}\right)[/tex]

Using the properties of logarithm, we get

[tex]\log_84+\log_8a+\log_8\left(\dfrac{b-4}{c^4}\right)[/tex]     [tex][\because \log_a mn=\log_a m+\log_a n][/tex]

[tex]\log_84+\log_8a+\log_8(b-4)-\log_8c^4[/tex]     [tex][\because \log_a \frac{m}{n}=\log_a m-\log_a n][/tex]

[tex]\log_84+\log_8a+\log_8(b-4)-4\log_8c[/tex]     [tex][\because \log_a x^n =n\log_a x][/tex]

Therefore, the required expression is [tex]\log_84+\log_8a+\log_8(b-4)-4\log_8c[/tex].

Answer:

B on edge

Step-by-step explanation:

e
65. the perpendicular
bisector of the
segment with
endpoints (-5/2,-2)
and (3, 5)
HELP PLEASE! Picture included!

Answers

Answer:

  44x +56y = 95

Step-by-step explanation:

To write the equation of the perpendicular bisector, we need to know the midpoint and we need to know the differences of the coordinates.

The midpoint is the average of the coordinate values:

  ((-2.5, -2) +(3, 5))/2 = (0.5, 3)/2 = (0.25, 1.5) = (h, k)

The differences of the coordinates are ...

  (3, 5) -(-2.5, -2) = (3 -(-2.5), 5 -(-2)) = (5.5, 7) = (Δx, Δy)

Then the perpendicular bisector equation can be written ...

  Δx(x -h) +Δy(y -k) = 0

  5.5(x -0.25) +7(y -1.5) = 0

  5.5x -1.375 +7y -10.5 = 0

Multiplying by 8 and subtracting the constant, we get ...

  44x +56y = 95 . . . . equation of the perpendicular bisector

The base of a rectangular prism is 20 cm 2. If the volume of the prism is 100 cm 3, what is its height?

Answers

Answer:

Step-by-step explanation:

Answer:

height = 5

Step-by-step explanation:

The volume of a prism is V = l*w*h

You are not given any information about the exact values of l and w.

You do know however that L and w when multiplied together = 20, so you can put that in for l*w. Then the formula becomes

V = 20*h

You are told that the volume is 100. Now the problem is simplified. You get

100 = 20 * h               Divide both sides by 20

100/20 = 20*h/20     Combine like terms.

5 = h

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