A random two digit number (10-99) is drawn. Find P(odd number)

Answer:

Step-by-step explanation:

[tex]100 (0.96)^{25} =[/tex] around 36.04

Answer:

220

Step-by-step explanation:

If we lost 12% we still have 100 - 12 = 88% of the employees left. 88% can be written as 0.88. 0.88 * 250 = 220 employees left.

Answer:

The number of computer is 5 and printer is 3

Answer:

The best estimate of the mean of the population is 50,000 miles, which is the sample mean.

To make a better inference, we know that the 95% confidence interval for the mean is (49,306; 50,694).

Step-by-step explanation:

The unbiased point estimation for the population mean tread life is the sample mean (50,000 miles), as it is the only information we have.

Although, knowing the standard deviation, we can calculate a confidence interval to make a stronger inference.

We calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=50000.

The sample size is N=100.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

[tex]s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3500}{\sqrt{100}}=\dfrac{3500}{10}=350[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=100-1=99[/tex]

The t-value for a 95% confidence interval and 99 degrees of freedom is t=1.98.

The margin of error (MOE) can be calculated as:

[tex]MOE=t\cdot s_M=1.98 \cdot 350=694.48[/tex]

Then, the lower and upper bounds of the confidence interval are:

[tex]LL=M-t \cdot s_M = 50000-694.48=49306\\\\UL=M+t \cdot s_M = 50000+694.48=50694[/tex]

The 95% confidence interval for the mean is (49306, 50694).

Properties of the logarithm: for any base of logarithm,

log(a*b) = log(a) + log(b)

If we replace b with 1/b, or b^-1, we have

log(a/b) = log(a) + log(1/b) = log(a) - log(b)

since

log(1/b) = log(b^-1) = - log(b)

using the power property of logarithms,

log(b^n) = n log(b)

Now,

ln35 = ln(5*7) = ln5 + ln7

ln(1/7) = - ln7

ln25 = ln(5^2) = 2 ln5

Putting everything together, we have

(ln35 + ln(1/7))/ln25 = (ln5 + ln7 - ln7)/(2 ln5) = ln5/(2 ln5) = 1/2

Answer:

4 hours.

Step-by-step explanation:

Well we can simply divide 32 by 8 and we get 4 hours:

32 miles ÷ 8 miles = 4 hours

It takes the cyclist 4 hours.

Answer:

  4 hours

Step-by-step explanation:

As every speed limit sign tells you, ...

  speed = miles/hours

Solving for time and using generic distance units, we get ...

  time = distance/speed

Filling in the given values, we have ...

  time = 32 km/(8 km/h) = (32/8) h = 4 h

It takes the cyclist 4 hours to travel 32 km.

Answer:

40% of 160 is 64

Step-by-step explanation:

You can easily find the answer in one step, just multiplying the whole (160) by the percentage (40) divided by 100.

So, 40% of 160 = 160 × 0.4 = 64.

Answer:

64

Step-by-step explanation:

You first have to subtract 40% from 160 and then you subtract that amount with is 96 from 160 and you get your answer 64

Answer: b) 45"

Step-by-step explanation:

Tv's are measured by their diagonal length.

Use Pythagorean Theorem to find the diagonal of the tv.

a² + b² = c²

36² + 27² = c²

1296 + 729 = c²

2025 = c²

√2025 = c

45 = c

Answer:

1/25

Step-by-step explanation:

5^-2

We know that a^ -b = 1/ a^b

5^-2 = 1/ 5^2

       = 1/25

Answer:

(a) [tex]\mathbf{P_2 = \left[\begin{array}{c}140 \\ 160 \\ 200 \end{array}\right]}[/tex]

(b)   After an infinite period of time; we will get back to a result similar to after the two time period which will be [tex]= \left[\begin{array}{c}140 \\ 160 \\ 200 \end{array}\right]}[/tex]

Step-by-step explanation:

The Markov Matrix can be interpret as :

[tex]M = \left[\begin{array}{ccc} \dfrac{1}{5} & \dfrac{2}{5} &\dfrac{1}{5} \\ \\ \dfrac{2}{5}&\dfrac{2}{5}&\dfrac{1}{5}\\ \\ \dfrac{2}{5}& \dfrac{2}{5}& \dfrac{2}{5} \end{array}\right][/tex]

From (a) ; we see that the initial population are as follows: 130 individuals in location 1, 300 in location 2, and 70 in location 3.

Le P represent the Population; So ;  [tex]P = \left[\begin{array}{c}130 \\ 300 \\ 70 \end{array}\right][/tex]

The objective is to find How many are in each location after two time periods;

So, after two time period ; we have the population [tex]P_2 = [M]^2 [P][/tex]

where;

[tex][M]^ 2 = \left[\begin{array}{ccc} \dfrac{1}{5} & \dfrac{2}{5} &\dfrac{1}{5} \\ \\ \dfrac{2}{5}&\dfrac{2}{5}&\dfrac{1}{5}\\ \\ \dfrac{2}{5}& \dfrac{2}{5}& \dfrac{2}{5} \end{array}\right] \left[\begin{array}{ccc} \dfrac{1}{5} & \dfrac{2}{5} &\dfrac{1}{5} \\ \\ \dfrac{2}{5}&\dfrac{2}{5}&\dfrac{1}{5}\\ \\ \dfrac{2}{5}& \dfrac{2}{5}& \dfrac{2}{5} \end{array}\right][/tex]

[tex][M]^ 2 = \dfrac{1}{25} \left[\begin{array}{ccc} 1+2+4 & 1+2+4 &1+2+4 \\ \\ 2+2+4&2+2+4&2+2+4\\ \\ 2+4+4&2+4+4& 2+4+4 \end{array}\right][/tex]

[tex][M]^ 2 = \dfrac{1}{25} \left[\begin{array}{ccc}7&7&7 \\ \\ 8 &8&8\\ \\10&10& 10 \end{array}\right][/tex]

Now; Over to after two time period ; when the population [tex]P_2 = [M]^2 [P][/tex]

[tex]P_2 = \dfrac{1}{25} \left[\begin{array}{ccc}7&7&7 \\ \\ 8 &8&8\\ \\10&10& 10 \end{array}\right] \left[\begin{array}{c}130 \\ 300 \\ 70 \end{array}\right][/tex]

[tex]\mathbf{P_2 = \left[\begin{array}{c}140 \\ 160 \\ 200 \end{array}\right]}[/tex]

(b) The total number of individuals in the migration process is 500. After a long time, how many are in each location?

After a long time; that is referring to an infinite time (n)

So; [tex]P_n = [M]^n [P][/tex]

where ;

[tex][M]^n \ can \ be \ [M]^2 , [M]^3 , [M]^4 .... \infty[/tex]

; if we determine the respective values of [tex][M]^2 , [M]^3 , [M]^4 .... \infty[/tex] we will always result to the value for [tex][M]^n[/tex]; Now if  [tex][M]^n[/tex] is said to be a positive integer; then :

After an infinite period of time; we will get back to a result similar to after the two time period which will be [tex]= \left[\begin{array}{c}140 \\ 160 \\ 200 \end{array}\right]}[/tex]

Answer:

[tex] P(0.34 <\hat p<0.49)[/tex]

And the distribution for the sample proportion is given by;

[tex]\hat p \sim N(p, \sqrt{\frac{p(1-p)}{n}})[/tex]

And we can find the mean and deviation for the sample proportion:

[te]\mu_{\hat p}= 0.36[/tex]

[tex]\sigma_{\hat p} =\sqrt{\frac{0.36(1-0.36)}{195}}= 0.0344[/tex]

And we can use the z score formula given by:

[tex] z = \frac{0.34 -0.36}{0.0344}= -0.582[/tex]

[tex] z = \frac{0.49 -0.36}{0.0344}= 3.782[/tex]

And we can use the normal distribution table and we got:

[tex] P(-0.582 <z< 3.782) =P(z<3.782)-P(z<-0.582)=0.99992-0.2803= 0.71962[/tex]

Step-by-step explanation:

For this case we know that the sample size is n =195 and the probability of success is p=0.36.

We want to find the following probability:

[tex] P(0.34 <\hat p<0.49)[/tex]

And the distribution for the sample proportion is given by;

[tex]\hat p \sim N(p, \sqrt{\frac{p(1-p)}{n}})[/tex]

And we can find the mean and deviation for the sample proportion:

[tex]\mu_{\hat p}= 0.36[/tex]

[tex]\sigma_{\hat p} =\sqrt{\frac{0.36(1-0.36)}{195}}= 0.0344[/tex]

And we can use the z score formula given by:

[tex] z = \frac{0.34 -0.36}{0.0344}= -0.582[/tex]

[tex] z = \frac{0.49 -0.36}{0.0344}= 3.782[/tex]

And we can use the normal distribution table and we got:

[tex] P(-0.582 <z< 3.782) =P(z<3.782)-P(z<-0.582)=0.99992-0.2803= 0.71962[/tex]

Answer:

y-intercept decreased by 3-1= 2 points

Step-by-step explanation:

y=-9x+ 3 ⇒ y-intercept= 3

y=-9x + 1 ⇒ y-intercept= 1

y-intercept decreased by 3-1= 2 points = the line shifted down by 2 points

Answer:

Hello!

Answer: y-intercept decreased by 2 points because 3-1=2

I hope I was of help.  If not, please let me know!  Thanks!

Step-by-step explanation:

Answer:

The number of students who scored within 2 standard deviations of the mean is 380.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

The number of students who scored within 2  standard deviations of the mean is approximately

By the Empirical Rule, 95% of the students scored within 2 standard deviations of the mean

Out of 400

0.95*400 = 380

The number of students who scored within 2 standard deviations of the mean is 380.

Answer: 380

Step-by-step explanation: 95% of 400 is 380

Area = length x width

Area = 5/6 x 3/4

= (5x3) / (6x4)

= 15/24

= 5/8 square miles

Answer:

x: -5,  -3,   0,  1,  4

y:-23, -15, -3, 1, 13        for the function.

x:-23, -15, -3, 1, 13

y: -5,   -3,  0,  1,  4         for the inverse.

Step-by-step explanation:

we know that if we have the function f(x) = y, then the inverse of f(x) (let's call it g(x)) is such that:

g(y) = x.

now we have

y=4x-3

y=(1/4)x+3/4

The only table that works for our first function is:

x: -5,  -3,   0,  1,  4

y:-23, -15, -3, 1, 13

You can see this by replacing the values of x and see if the value of y also coincides.

Then, using the fact that the other table must be for the inverse, we should se a table with the same values, but where the values of x and y are interchanged.

The second table is that one:

x:-23, -15, -3, 1, 13

y: -5,   -3,  0,  1,  4

Answer: B and D

x:-23, -15, -3, 1, 13

y:-5, -3, 0, 1, 4

x:-5, -3, 0, 1, 4

y:-23, -15, -3, 1, 13

Step-by-step explanation:

Answer:

On average it will take 13 hrs 53 minutes before the van is filled

Step-by-step explanation:

The first thing we need to do here is to find find the number of defective light bulbs

Using the poisson process, that would be;

λ * p

where λ is the poisson rate of production which is 300 per hour

and p is the probability that the produced bulb is defective = 0.012

So the number of defective bulbs produced within the hour = 0.012 * 300 = 3.6 light bulbs per hour

Now, let X be the time until 50 light bulbs are produced. Then X is a random variable with the parameter (r, λ) = (50, 3.6)

What we need to find however is E(X)

Thus, the expected value of a gamma random variable X with the parameter (x, λ) is;

E(X) = r/λ = 50/3.6 = 13.89

Thus the amount of time it will take before the Can will be filled is 13 hrs 53 minutes

Answer:

P(Y= 0) = 0.1

P(Y= 0) = 0.7

P(Y= 0) = 0.2

Step-by-step explanation:

Let Y be number of impurities that can be found in the well,

Let A denote the event that impurity A is randomly found in the well

Here Y can have three values i.e 0 , 1 and 2

✓It will take take the value of 0 when there is no impurity found in the well

✓It will take the value of 1 when when exactly one impurity vis found in the well

✓It will take the value of 2 when when both impurities vis found in the well

CHECK THE ATTACHMENT FOR DETAILED EXPLATION

Answer:person up top is right it’s B

Step-by-step explanation: on edg 2020

Answer:

The answer is B

Step-by-step explanation:

lol yw guys

Step-by-step explanation:

a rectangle has reflectional symmetry when reflected over the line through the midpoints of its opposite sides

Answer: A square always has a four reflection symmetry no matter the size.

Answer:

√130 is the distance between (9,-3) (0,-10)

Answer:

x=3

Step-by-step explanation:

4x^2 - 36 = 0

Add 36 to each side

4x^2 -36 +36 = 0+36

4x^2 = 36

Divide each side by 4

4x^2/4 =36/4

x^2 = 9

Take the square root of each sdie

sqrt(x^2) = ±sqrt(3)

x = -3,+3

We want the positive square root

x=3

Answer:

Step-by-step explanation:

Answer:

979

Step-by-step explanation:

Answer:

115/198

Step-by-step explanation:

khan

Answer:

lily has a larger ratio

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

(eˣ − e⁻ˣ) / (eˣ + e⁻ˣ) = t

Multiply by eˣ/eˣ.

(e²ˣ − 1) / (e²ˣ + 1) = t

Solve for e²ˣ.

e²ˣ − 1 = (e²ˣ + 1) t

e²ˣ − 1 = e²ˣ t + t

e²ˣ = 1 + e²ˣ t + t

e²ˣ − e²ˣ t = 1 + t

e²ˣ (1 − t) = 1 + t

e²ˣ = (1 + t) / (1 − t)

Solve for x.

2x = ln[(1 + t) / (1 − t)]

x = ½ ln[(1 + t) / (1 − t)]

Use log rule.

x = ln(√[(1 + t) / (1 − t)])

Answer:

7.33333333333 I think. Hope this helped.

Answer:

  (-2.60, -6.80)

Step-by-step explanation:

The new coordinates can be found by multiplying by the rotation matrix:

  [tex]\left[\begin{array}{c}x'&y'\end{array}\right]=\left[\begin{array}{cc}\cos{\theta}&-\sin{\theta}\\\sin{\theta}&\cos{\theta}\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right][/tex]

That is, ...

  x' = x·cos(175°) -y·sin(175°) = 2(-0.9962) -7(0.0872) = -2.60

  y' = x·sin(175°) +y·cos(175°) = 2(0.0872) +7(-0.9962) = -6.80

The new coordinates are ...

  (x', y') = (-2.60, -6.80)

Answer:

Step-by-step explanation:

Corresponding gasoline consumption when radial tires is used and gasoline consumption when regular belted tires is used form matched pairs.

The data for the test are the differences between the gasoline consumption when radial tires is used and gasoline consumption when regular belted tires is used.

μd = the gasoline consumption when radial tires is used minus the gasoline consumption when regular belted tires is used.

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The resulting p-value was .0152.

Since alpha, 0.05 > than the p value, 0.0152, then we would reject the null hypothesis. Therefore, at 5% significance level, we can conclude that the gasoline consumption when regular belted tires is used is higher than the gasoline consumption when radial tires is used.

Answer:

-$0.75

Step-by-step explanation:

For calculation of expected value first we need to find out the probability distribution for this raffle which is shown below:-

Amount                  Probability

500 - 1 = $499       1 ÷ 5,000

200 - 1 = $199        3 ÷ 5,000

10 - 1 = $9                5 ÷ 5,000

5 - 1 = $4                   20 ÷ 5,000

-$1                             5,000 - 29 ÷ 5,000 = 4,971 ÷ 5,000

Now, the expected value of raffle will be

[tex]= \$499 \times (\frac{1}{5,000}) + \$199 \times (\frac{3}{5,000}) + \$9 \times (\frac{5}{5,000}) + \$4 \times (\frac{20}{5,000}) - \$1 \times (\frac{4,971}{5,000})[/tex]

= 0.0998  + 0.1194  + 0.009  + 0.016  - 0.9942

= -$0.75

The expected value of this raffle per ticket is $ 0.25.

Given that five thousand tickets are sold at $ 1 each for a charity raffle, and tickets are to be drawn at random and monetary prizes awarded as follows: 1 prize of $ 500, 3 prizes of $ 200, 5 prizes of $ 10, and 20 prizes of $ 5, to determine what is the expected value of this raffle if you buy 1 ticket, the following calculation must be performed:

Therefore, the expected value of this raffle per ticket is $ 0.25.

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