Answer:
Yes, we have reason to believe that fewer than one-fifth are heated by oil.
Step-by-step explanation:
A one-sample proportion test is to be performed to determine whether fewer than one-fifth of the homes in a certain city are heated by oil.
The hypothesis can be defined as follows:
H₀: The proportion of homes in a certain city that are heated by oil is not less than one-fifth, i.e. p ≥ 0.20.
Hₐ: The proportion of homes in a certain city that are heated by oil is less than one-fifth, i.e. p < 0.20.
The information provided is:
n = 1000
x = 136
α = 0.05
Compute the sample proportion as follows:
[tex]\hat p=\frac{x}{n}=\frac{136}{1000}=0.136[/tex]
Compute the test statistic as follows:
[tex]z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}[/tex]
[tex]=\frac{0.136-0.20}{\sqrt{\frac{0.136(1-0.136)}{1000}}}\\\\=-5.9041\\\\\approx -5.90[/tex]
The test statistic value is, -5.90.
Decision rule:
Reject the null hypothesis if the p-value of the test is less than the significance level.
Compute the p-value as follows:
[tex]p-value=P(Z<-5.90)\\\\=1-P(Z<5.90)\\\\=1-(\approx 1)\\\\=0[/tex]
The p-value of the test is, 0.
p-value = 0 < α = 0.05
The null hypothesis will be rejected at 5% level of significance.
Conclusion:
The proportion of homes in a certain city that are heated by oil is less than one-fifth.
What’s the correct answer for this question?
Answer:
C
Step-by-step explanation:
A cylinder is formed when rotating the 3-D figure around y-axis
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
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.
A student takes a multiple-choice test that has 11 questions. Each question has five choices. The student guesses randomly at each answer. Let X be the number of questions answered correctly. (a) Find P (6). (b) Find P (More than 3). Round the answers to at least four decimal places.
Answer:
a) P(6) = 0.0097
b) P(More than 3) = 0.1611
Step-by-step explanation:
For each question, there are only two possible outcomes. Either it is guessed correctly, or it is not. Questions are independent of each other. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
A student takes a multiple-choice test that has 11 questions.
This means that [tex]n = 11[/tex]
Each question has five choices.
This means that [tex]p = \frac{1}{5} = 0.2[/tex]
(a) Find P (6)
This is P(X = 6).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 6) = C_{11,6}.(0.2)^{6}.(0.8)^{5} = 0.0097[/tex]
P(6) = 0.0097
(b) Find P (More than 3).
Either P is 3 or less, or it is more than three. The sum of the probabilities of these outcomes is 1. So
[tex]P(X \leq 3) + P(X > 3) = 1[/tex]
We want P(X > 3). So
[tex]P(X > 3) = 1 - P(X \leq 3)[/tex]
In which
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{11,0}.(0.2)^{0}.(0.8)^{11} = 0.0859[/tex]
[tex]P(X = 1) = C_{11,1}.(0.2)^{1}.(0.8)^{10} = 0.2362[/tex]
[tex]P(X = 2) = C_{11,2}.(0.2)^{2}.(0.8)^{9} = 0.2953[/tex]
[tex]P(X = 3) = C_{11,3}.(0.2)^{3}.(0.8)^{8} = 0.2215[/tex]
[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0859 + 0.2362 + 0.2953 + 0.2215 = 0.8389[/tex]
Then
[tex]P(X > 3) = 1 - P(X \leq 3) = 1 - 0.8389 = 0.1611[/tex]
P(More than 3) = 0.1611
Find the area of a circle with radius, r = 5.7m.
Give your answer rounded to 2 DP.
The diagram is not drawn to scale.
(I attached the diagram below!)
Answer:
the area of the circle is 102.11 square metres
How can you use mathematics to help scientists explore Martian Craters ?
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.
11+11=4
22+22=16
33+33=
What’s the answer
Answer:
what method exactly r u using ????
Which expression is equivalent to log Subscript 8 Baseline 4 a (StartFraction b minus 4 Over c Superscript 4 Baseline EndFraction)?
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:
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________.
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 = 9So, 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.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.
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
One number is 4 plus one half of another number. Their sum is 31. Find the numbers.
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
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
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
e
65. the perpendicular
bisector of the
segment with
endpoints (-5/2,-2)
and (3, 5)
HELP PLEASE! Picture included!
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
Graph the line with slope -1/3 and y -intercept 6 .
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:
m is directly proportional to r squared when r=2 m=14 work out the value of r when m = 224
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:
The base of a rectangular prism is 20 cm 2. If the volume of the prism is 100 cm 3, what is its height?
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
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.
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
which is pattern 12,24,36,48
Answer:
multiples of 12
Step-by-step explanation: when looking at the GCF, the answer is 12
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.
Answer:
work is shown and pictured
I need help with this one
Answer:
2 2/3
Step-by-step explanation:
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
Answer:
The correct answer is B (24 units)
Step-by-step explanation:
what is the common ratio of the geometric sequence below ?
-96,48,-24,12,-6...
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:
Write an
explicit formula for
ans
the nth
term of the sequence 20, -10,5, ....
Answer:an=20(-1/2)^n-1
Step-by-step explanation:
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
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
What is the center of the circle?
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...
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
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
Any help would be great
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%
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
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
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
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]
arl rides his bicycle 120 feet in 10 seconds. How many feet does he ride in 1 minute? 2 feet 12 feet 720 feet 7,200 feet
Answer: 720 ft
Step-by-step explanation: He rides 720 feet.
if 120 feet are in 10 seconds then;
60 seconds are 60/10*120=720 feet
Answer:
720
Step-by-step explanation:
120/10 to find his feet per second which is 12 feet per second
12*60
since there are 60 seconds in a minute
= 720
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
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
