Renee is making a scale diagram of her MP3 player. The length of her scale drawing is 8 inches, and the width is 14 inches. The actual length of the MP3 player is 4 centimeters, and the width is 7 centimeters. This is , and the scale factor is .

Answer:

I'm not completely sure what you mean by a, "single fraction," but I'm pretty sure the answer you are looking for is [tex]\frac{5-4}{a-b}[/tex]

Step-by-step explanation:

Answer:

[tex]z=-\frac{20}{3}[/tex]

Step-by-step explanation:

[tex]\frac{3z}{10}-4=-6\\\\\frac{3z}{10}-4+4=-6+4\\\\\frac{3z}{10}=-2\\\\\frac{10\cdot \:3z}{10}=10\left(-2\right)\\\\3z=-20\\\\\frac{3z}{3}=\frac{-20}{3}\\\\z=-\frac{20}{3}[/tex]

Best Regards!

Answer:

1)

A) [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B) P([tex]\frac{}{X}[/tex] > 9)= 0.0552

C) P(X> 9)= 0.36317

D) IQR= 0.4422

2)

A) [tex]\frac{}{X}[/tex] ~ N(30;2.5)

B) P( [tex]\frac{}{X}[/tex]<30)= 0.50

C) P₉₅= 32.60

D) P( [tex]\frac{}{X}[/tex]>36)= 0

E) Q₃: 31.0586

Step-by-step explanation:

Hello!

1)

The variable of interest is

X: pollutants found in waterways near a large city. (ppm)

This variable has a normal distribution:

X~N(μ;σ²)

μ= 8.5 ppm

σ= 1.4 ppm

A sample of 18 large cities were studied.

A) The sample mean is also a random variable and it has the same distribution as the population of origin with exception that it's variance is affected by the sample size:

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

The population mean is the same as the mean of the variable

μ= 8.5 ppm

The standard deviation is

σ/√n= 1.4/√18= 0.329= 0.33 ⇒σ²/n= 0.33²= 0.108

So: [tex]\frac{}{X}[/tex] ~ N(8.5;0.108)

B)

P([tex]\frac{}{X}[/tex] > 9)= 1 - P([tex]\frac{}{X}[/tex] ≤ 9)

To calculate this probability you have to standardize the value of the sample mean and then use the Z-tables to reach the corresponding value of probability.

Z= [tex]\frac{\frac{}{X} - Mu}{\frac{Sigma}{\sqrt{n} } } = \frac{9-8.5}{0.33}= 1.51[/tex]

Then using the Z table you'll find the probability of

P(Z≤1.51)= 0.93448

Then

1 - P([tex]\frac{}{X}[/tex] ≤ 9)= 1 - P(Z≤1.51)= 1 - 0.93448= 0.0552

C)

In this item, since only one city is chosen at random, instead of working with the distribution of the sample mean, you have to work with the distribution of the variable X:

P(X> 9)= 1 - P(X ≤ 9)

Z= (X-μ)/δ= (9-8.5)/1.44

Z= 0.347= 0.35

P(Z≤0.35)= 0.63683

Then

P(X> 9)= 1 - P(X ≤ 9)= 1 - P(Z≤0.35)= 1 - 0.63683= 0.36317

D)

The first quartile is the value of the distribution that separates the bottom 2% of the distribution from the top 75%, in this case it will be the value of the sample average that marks the bottom 25% symbolically:

Q₁: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₁)= 0.25

Which is equivalent to the first quartile of the standard normal distribution. So first you have to identify the first quartile for the Z dist:

P(Z≤z₁)= 0.25

Using the table you have to identify the value of Z that accumulates 0.25 of probability:

z₁= -0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₁= ([tex]\frac{}{X}[/tex]₁-μ)/(σ/√n)

z₁*(σ/√n)= ([tex]\frac{}{X}[/tex]₁-μ)

[tex]\frac{}{X}[/tex]₁= z₁*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₁= (-0.67*0.33)+8.5=  8.2789 ppm

The third quartile is the value that separates the bottom 75% of the distribution from the top 25%. For this distribution, it will be that value of the sample mean that accumulates 75%:

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

Using the table you have to identify the value of Z that accumulates 0.75 of probability:

z₃= 0.67

Now you have to translate the value of Z to a value of [tex]\frac{}{X}[/tex]:

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*0.33)+8.5=  8.7211 ppm

IQR= Q₃-Q₁= 8.7211-8.2789= 0.4422

2)

A)

X ~ N(30,10)

For n=4

[tex]\frac{}{X}[/tex] ~ N(μ;σ²/n)

Population mean μ= 30

Population variance σ²/n= 10/4= 2.5

Population standard deviation σ/√n= √2.5= 1.58

[tex]\frac{}{X}[/tex] ~ N(30;2.5)

B)

P( [tex]\frac{}{X}[/tex]<30)

First you have to standardize the value and then look for the probability:

Z=  ([tex]\frac{}{X}[/tex]-μ)/(σ/√n)= (30-30)/1.58= 0

P(Z<0)= 0.50

Then

P( [tex]\frac{}{X}[/tex]<30)= 0.50

Which is no surprise since 30 y the value of the mean of the distribution.

C)

P( [tex]\frac{}{X}[/tex]≤ [tex]\frac{}{X}[/tex]₀)= 0.95

P( Z≤ z₀)= 0.95

z₀= 1.645

Now you have to reverse the standardization:

z₀= ([tex]\frac{}{X}[/tex]₀-μ)/(σ/√n)

z₀*(σ/√n)= ([tex]\frac{}{X}[/tex]₀-μ)

[tex]\frac{}{X}[/tex]₀= z₀*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₀= (1.645*1.58)+30= 32.60

P₉₅= 32.60

D)

P( [tex]\frac{}{X}[/tex]>36)= 1 - P( [tex]\frac{}{X}[/tex]≤36)= 1 - P(Z≤(36-30)/1.58)= 1 - P(Z≤3.79)= 1 - 1 = 0

E)

Q₃: P([tex]\frac{}{X}[/tex]≤[tex]\frac{}{X}[/tex]₃)= 0.75

⇒ P(Z≤z₃)= 0.75

z₃= 0.67

z₃= ([tex]\frac{}{X}[/tex]₃-μ)/(σ/√n)

z₃*(σ/√n)= ([tex]\frac{}{X}[/tex]₃-μ)

[tex]\frac{}{X}[/tex]₃= z₃*(σ/√n)+μ

[tex]\frac{}{X}[/tex]₃= (0.67*1.58)+30= 31.0586

Q₃: 31.0586

Answer:

(a)The total sales of ice-cream last month is 702 gallons.

(b)The total sales of broccoli last month is 510 pounds.

Step-by-step explanation:

Part A

Total Sales of gallons of ice cream this month = 597

Since it represents a decrease of 15% of last month's sales

Let the total sales of ice-cream last month =x

Then:

(100-15)% of x =597

85% of x=597

0.85x=597

x=597/0.85

x=702 (to the nearest integer)

The total sales of ice-cream last month is 702 gallons.

Part B

Total Sales of broccoli this month = 617 pounds

Since it represents an increase of 21% of last month's sales

Let the total sales of ice-cream last month =y

Then:

(100+21)% of y =617

121% of y=617

1.21y=617

y=617/1.21

y=510 (to the nearest integer)

The total sales of broccoli last month is 510 pounds.

Answer:

95% for the true average increase in total cholesterol under these circumstances

(-2.306 , 9.406)

Step-by-step explanation:

Step(i):-

Given sample size 'n' =41

Mean of the sample(x⁻)  = 3.55

The estimated standard error

                                             [tex]S.E = \frac{S.D}{\sqrt{n} }[/tex]

Given  estimated standard error ( S.E) = 3.478

Level of significance ∝=0.05

Step(ii):-

95% for the true average increase in total cholesterol under these circumstances

[tex](x^{-} - t_{0.05} S.E ,x^{-} + t_{0.05} S.E)[/tex]

Degrees of freedom

ν= n-1 = 41-1 =40

t₀.₀₅ = 1.6839

95% for the true average increase in total cholesterol under these circumstances

[tex](x^{-} - t_{0.05} S.E ,x^{-} + t_{0.05} S.E)[/tex]

( 3.55 - 1.6839 ×3.478 ,3.55 + 1.6839 ×3.478 )

(3.55 - 5.856 , 3.55 + 5.856)

(-2.306 , 9.406)

Conclusion:-

95% for the true average increase in total cholesterol under these circumstances

(-2.306 , 9.406)

Answer:

The value of the sample mean resonance frequency is 112Hz

Step-by-step explanation:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

In this problem, we have that:

Lower bound: 111.6

Upper bound: 112.4

Sample mean: (111.6 + 112.4)/2 = 112Hz

The value of the sample mean resonance frequency is 112Hz

The value of the sample mean resonance frequency is 112 Hz.

The value of the sample mean resonance frequency is equivalent to the average of the upper limit and the lower limit.

The sample mean resonance frequency = (lower limit + upper limit) / 2

(111.6 +112.4) / 2

= 224 / 2

= 112 Hz

To learn more about confidence interval, please check: https://brainly.com/question/15905477

Answer:

10 and 8

Step-by-step explanation:

10 and 8 are the factors in this equation because factors are the numbers that are mutiplied together to get the product (The answer to a mutiplication problem) Therefore the factors in this equation are 10 and 8 because those are the numbers that are mutiplied together to get the product.

Answer:

Is possible to make a Type I error, where we reject a true null hypothesis.

Step-by-step explanation:

We have a hypothesis test of a proportion. The claim is that the proportion of paid leave has significantly decrease from 2011 to january 2012. The P-value for this test is 0.0271 and the nunll hypothesis is rejected.

As the conclusion is to reject the null hypothesis, the only error that we may have comitted is rejecting a true null hypothesis.

The null hypothesis would have stated that there is no significant decrease in the proportion of paid leave.

This is a Type I error, where we reject a true null hypothesis.

Answer:

The third options

Step-by-step explanation:

Counting we can see that 10 students went to two or less states, and 10 went to three or more

Answer:

The probability that demand during lead-time will exceed 20 bails is 0.2033.

Step-by-step explanation:

We are given that it has been previously determined that demand during the lead-time is normally distributed with a mean of 15 bails and a standard deviation of 6 bails.

Let X = demand during the lead-time

So, X ~ Normal([tex]\mu=15, \sigma^{2} = 6^{2}[/tex])

The z-score probability distribution for the normal distribution is given by;

                               Z  =  [tex]\frac{X-\mu}{\sigma}[/tex]  ~ N(0,1)

where, [tex]\mu=[/tex] population mean demand = 15 bails

           [tex]\sigma[/tex] = standard deviation = 6 bails

Now, the probability that demand during lead-time will exceed 20 bails is given by = P(X > 20 bails)

       P(X > 20 bails) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{20-15}{6}[/tex] ) = P(Z > 0.83) = 1 - P(Z [tex]\leq[/tex] 0.83)

                                                             = 1 - 0.7967 = 0.2033

Answer:

Question 1: 3 - 5 hours.

Question 2: 0 - 1 hour

Step-by-step explanation:

Question 1: As you can see in the diagram, the guy is moving really slowly and is almost stuck, therefore, it is 3 - 5  hours.

Question 2:  In hours 0 - 1, you can see that the graph is the closest to vertical as it gets.

Answer:

root 61

Step-by-step explanation:

You can use the distance formula or draw a triangle with sides 5 and 6

Answer:

The answer is B

Step-by-step explanation:

Hope this helps.. if not im sorry :(

Answer/Step-by-step explanation:

When solving problems like this, remember the following:

1. + × + = +

2. + × - = -

3. - × + = -

4. - × - = +

Let's solve:

a. (-4) + (+10) + (+4) + (-2)

Open the bracket

- 4 + 10 + 4 - 2

= - 4 - 2 + 10 + 4

= - 6 + 14 = 8

b. (+5) + (-8) + (+3) + (-7)

= + 5 - 8 + 3 - 7

= 5 + 3 - 8 - 7

= 8 - 15

= - 7

c. (-19) + (+14) + (+21) + (-23)

= - 19 + 14 + 21 - 23

= - 19 - 23 + 14 + 21

= - 42 + 35

= - 7

d. (+5) - (-10) - (+4)

= + 5 + 10 - 4

= 15 - 4 = 11

e. (-3) - (-3) - (-3)

= - 3 + 3 + 3

= - 3 + 9

= 6

f. (+26) - (-32) - (+15) - (-8)

= 26 + 32 - 15 + 8

= 26 + 32 + 8 - 15

= 66 - 15

= 51

Answer:

No solutions.

Step-by-step explanation:

9|x-1|=-45

Divide 9 into both sides.

|x-1| = -45/9

|x-1| = -5

Absolute value cannot be less than 0.

Answer:

No solution

Step-by-step explanation:

=> 9|x-1| = -45

Dividing both sides by 9

=> |x-1| = -5

Since, this is less than zero, hence the equation has no solutions

Answer:

Step-by-step explanation:

given a point [tex](x_0,y_0)[/tex] the equation of a line with slope m that passes through the  given point is

[tex]y-y_0 = m(x-x_0)[/tex] or equivalently

[tex] y = mx+(y_0-mx_0)[/tex].

Recall that a line of the form [tex]y=mx+b [/tex], the y intercept is b and the x intercept is [tex]\frac{-b}{m}[/tex].

So, in our case, the y intercept is [tex](y_0-mx_0)[/tex] and the x  intercept is [tex]\frac{mx_0-y_0}{m}[/tex].

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph [tex](x_0,\frac{1}{x_0})[/tex]. Which means that [tex]y_0=\frac{1}{x_0}[/tex]

The slope of the tangent line is given by the derivative of the function evaluated at [tex]x_0[/tex]. Using the properties of derivatives, we get

[tex]y' = \frac{-1}{x^2}[/tex]. So evaluated at [tex]x_0[/tex] we get [tex] m = \frac{-1}{x_0^2}[/tex]

Replacing the values in our previous findings we get that the y intercept is

[tex](y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}[/tex]

The x intercept is

[tex] \frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0[/tex]

The triangle in consideration has height [tex]\frac{2}{x_0}[/tex] and base [tex]2x_0[/tex]. So the area is

[tex] \frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2[/tex]

So regardless of the point we take on the graph, the area of the triangle is always 2.

Answer:

41 word/min

Step-by-step explanation:

Before noon Ali works:

She types:

After lunch she works:

She types:

Total Ali works= 4+4= 8 hours= 480 min

Total Ali types= 11520+8160= 19680 words

Average typing rate= 19680 words/480 min= 41 word/min

Answer:

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

Answer:

Step-by-step explanation:

From the information given, the population is divided into sub groups. Each group would consist of citizens picking a particular choice as the most important problem facing the country. The choices are the different categories. In this case, the null hypothesis would state that the distribution of proportions for all categories is the same in each population. The alternative hypothesis would state that the distributions is different. Therefore, the correct test to use to determine if the distribution of​ "problem facing this country ​today" is different between the two different​ years is

A) Use a​ chi-square test of homogeneity.

Answer: 3.9

Step-by-step explanation: Khan Academy

The length of BD if The angle ∠ DAC is equal to the angle ∠ BAD is 3.92.

Three straight lines coming together create a triangle. There are three sides and three corners on every triangle (angles). A triangle's vertex is the intersection of two of its sides. Any one of a triangle's three sides can serve as its base, however typically the bottom side is used.

Given:

The angle ∠ DAC = angle ∠ BAD

As we can see that the triangle BAD and triangle DAC are similar By SAS similarity,

AC / AB = CD / BD  (By the proportional theorem of similarity)

5.6 / 5.1 = 4.3 / BD

1.09 = 4.3 / BD

BD = 4.3 / 1.09

BD = 3.92

Thus, the length of BD is 3.92.

To know more about Triangles:

https://brainly.com/question/16886469

#SPJ2

Answer:

-30000/100

300 0/1

Step-by-step explanation:

We have the following numbers -18 and 16 2/3, the first is an integer and the second is a mixed number, the first thing is to pass the mixed number to a decimal number.

16 2/3 = 16.67

We do the multiplication:

−18⋅16 2/3 = -300

We have an improper fraction is a fraction in which the numerator (top number) is greater than or equal to the denominator (bottom number), therefore it would be:

-30000/100

How mixed number would it be:

300 0/1

Answer:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

Answer:

Step-by-step explanation:

[tex]\frac{5}{6}-\frac{2}{3}=\frac{5}{6}-\frac{2*2}{3*2}\\\\=\frac{5}{6}-\frac{4}{6}\\\\=\frac{5-4}{6}\\\\=\frac{1}{6}[/tex]

Answer:

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

[tex]M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\[/tex]

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu=500\\\\H_a:\mu< 500[/tex]

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

The estimated standard error of the mean is computed using the formula:

[tex]s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1[/tex]

The degrees of freedom for this sample size are:

[tex]df=n-1=5-1=4[/tex]

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=P(t<-1.1)=0.166[/tex]

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

Answer:

the answer is 3

Step-by-step explanation:

i took the test

Answer : The value of x is 4.1 cm.

Step-by-step explanation :

As we know that the perpendicular dropped from the center divides the chord into two equal parts.

That means,

AB = CB = [tex]\frac{15.6cm}{2}=7.8cm[/tex]

Now we have o calculate the value of x by using Pythagoras theorem.

Using Pythagoras theorem in ΔOBA :

[tex](Hypotenuse)^2=(Perpendicular)^2+(Base)^2[/tex]

[tex](OA)^2=(OB)^2+(BA)^2[/tex]

Now put all the values in the above expression, we get the value of side OB.

[tex](8.8)^2=(x)^2+(7.8)^2[/tex]

[tex]x=\sqrt{(8.8)^2-(7.8)^2}[/tex]

[tex]x=\sqrt{77.44-60.84}[/tex]

[tex]x=\sqrt{16.6}[/tex]

[tex]x=4.074\approx 4.1[/tex]

Therefore, the value of x is 4.1 cm.

Answer:

Just replace the variables with the number

d5

c4 (uh oh)

a2

b-3

f-7

d-c = 5 - 4 = 1

1/3 - 4(ab+f)

2 x -3 = -6

-6 + -7 = -13

-13 x 4 = -52

1/3 - -52 = 1/3 + 52 =

52 1/3

Hope this helps

Step-by-step explanation:

Answer:

Option C is correct.

The sampling distribution with sample size n=100 will have less variability.

Step-by-step explanation:

Complete Question

Consider random samples selected from the population of all female college soccer players in the United States. Assume the mean height of female college soccer players in the United States is 66 inches and the standard deviation is 3.5 inches. Which do you expect to have less variability (spread): a sampling distribution with sample size n = 100 or a sample size of n = 20.

A. Both sampling distributions will have the same variability.

B.The sampling distribution with sample size n=20 will have less variability

C. The sampling distribution with sample size n =100 will have less variability

Solution

The central limit theorem allows us to say that as long as

- the sample is randomly selected from the population distribution with each variable independent of each other and with the sample having an adequate enough sample size.

- the random sample is normal or almost normal which is guaranteed if the population distribution that the random sample was extracted from is normal or approximately normal,

1) The mean of sampling distribution (μₓ) is approximately equal to the population mean (μ)

μₓ = μ = 66 inches

2) The standard deviation of the sampling distribution or the standard error of the sample mean is related to the population standard deviation through

σₓ = (σ/√N)

where σ = population standard deviation = 3.5 inches

N = Sample size

And the measure of variability for a sampling distribution is the magnitude of the standard deviation of the sampling distribution.

For sampling distribution with sample size n = 100

σₓ = (3.5/√100) = 0.35 inch

For sampling distribution with sample size n = 20

σₓ = (3.5/√20) = 0.7826 inch

The standard deviation of the sampling distribution with sample size n = 20 is more than double that of the sampling distribution with sample size n = 100, hence, it is evident that the bigger the sample size, the lesser the standard deviation of the sampling distribution and the lesser the variability that the sampling distribution shows.

Hope this Helps!!!

Answer:

50,063,860 different samples can be chosen

Step-by-step explanation:

The order in which the microchips are chosen is not important. So we use the combinations formula to solve this question.

Combinations formula:

[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]

How many different samples can be chosen

We choose 6 microchips from a set of 60. So

[tex]C_{60,6} = \frac{60!}{6!(60-6)!} = 50063860[/tex]

50,063,860 different samples can be chosen

Answer:

y = 2/3x + 4

Step-by-step explanation:

Step 1: Find slope

m = (4-0)/(0+6)

m = 2/3

Step 2: Write in y-int (0, 4)

y = 2/3x + 4

Answer:

The numbers of doors that will have no blemishes will be about 6065 doors

Step-by-step explanation:

Let the number of counts by the  worker of each blemishes on the door be (X)

The distribution of blemishes followed the Poisson distribution with parameter  [tex]\lambda =0.5[/tex] / door

The probability mass function on of a poisson distribution Is:

[tex]P(X=x) = \dfrac{e^{- \lambda } \lambda ^x}{x!}[/tex]

[tex]P(X=x) = \dfrac{e^{- \ 0.5 }( 0.5)^ x}{x!}[/tex]

The probability that no blemishes occur is :

[tex]P(X=0) = \dfrac{e^{- \ 0.5 }( 0.5)^ 0}{0!}[/tex]

[tex]P(X=0) = 0.60653[/tex]

P(X=0) = 0.6065

Assume the number of paints on the door by q = 10000

Hence; the number of doors that have no blemishes  is = qp

=10,000(0.6065)

= 6065