can someone help me with this

Answer: 1/8

Step-by-step explanation: If you convert 1/4 to eighths, it becomes 2/8. 3/8-2/8 is 1/8

Answer:

The unit price is $5.

Answer:

$5.00.

Step-by-step explanation:

This is 22.5 / 4.5

= $5.00

Answer:

We need to sample at least 1069 parents.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.9}{2} = 0.05[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.05 = 0.95[/tex], so [tex]z = 1.645[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

How many parents do you have to sample?

We need to sample at least n parents.

n is found when [tex]M = 4, \sigma = 79.5[/tex]. So

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]4 = 1.645*\frac{79.5}{\sqrt{n}}[/tex]

[tex]4\sqrt{n} = 1.645*79.5[/tex]

[tex]\sqrt{n} = \frac{1.645*79.5}{4}[/tex]

[tex](\sqrt{n})^{2} = (\frac{1.645*79.5}{4})^{2}[/tex]

[tex]n = 1068.92[/tex]

Rounding up

We need to sample at least 1069 parents.

Answer:

[tex]n=(\frac{1.64(79.5)}{4})^2 =1062.43 \approx 1063[/tex]

So the answer for this case would be n=1063 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=79.5[/tex] represent the population standard deviation

n represent the sample size  

Solution to the problem

The margin of error is given by this formula:

[tex] ME=z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]    (a)

And on this case we have that ME =4 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

[tex]n=(\frac{z_{\alpha/2} \sigma}{ME})^2[/tex]   (b)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.05;0;1)", and we got [tex]z_{\alpha/2}=1.64[/tex], replacing into formula (b) we got:

[tex]n=(\frac{1.64(79.5)}{4})^2 =1062.43 \approx 1063[/tex]

So the answer for this case would be n=1063 rounded up to the nearest integer

Answer:

1) Null hypothesis:[tex]\mu \leq 500[/tex]  

Alternative hypothesis:[tex]\mu > 500[/tex]  

[tex]z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90[/tex]  

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

[tex]z_{crit}= 2.33[/tex]

For this case we see that the calculated value is higher than the critical value

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

2) Since is a right tailed test the p value would be:  

[tex]p_v =P(z>5.90)=1.82x10^{-9}[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

Step-by-step explanation:

Part 1

Data given

[tex]\bar X=640[/tex] represent the sample mean

[tex]\sigma=150[/tex] represent the population standard deviation

[tex]n=40[/tex] sample size  

[tex]\mu_o =500[/tex] represent the value that we want to test  

[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

[tex]p_v[/tex] represent the p value for the test (variable of interest)  

Step1:State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 500, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 500[/tex]  

Alternative hypothesis:[tex]\mu > 500[/tex]  

Step 2: Calculate the statistic

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)  

We can replace in formula (1) the info given like this:  

[tex]z=\frac{640-500}{\frac{150}{\sqrt{40}}}=5.90[/tex]  

Step 3: Calculate the critical value

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.01 of the area in the right and we got:

[tex]z_{crit}= 2.33[/tex]

Step 4: Compare the statistic with the critical value

For this case we see that the calculated value is higher than the critical value

Step 5: Decision

Since the calculated value is higher than the critical value we have enugh evidence to reject the null hypothesis at 1% of significance level

Part 2

P-value  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>5.90)=1.82x10^{-9}[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, same conclusion for part 1

The height of the shrubs is modeled by the function h(t) = 0.65t² + 2t + 17. After 7 years, the shrubs are 52.85 cm tall when they are sold.

The height (h) of a shrub after t years can be found by integrating the given growth rate function dh/dt = 1.3t + 2 with respect to time t. Given the initial height (or the initial condition), h(0) = 17 cm, the function turns out to be an integral taking initial condition into account:

h(t) = ∫ (1.3t + 2) dt + h(0)

After performing the integration, the function for height turns out to be:

[tex]h(t) = 0.65t^2 + 2t + 17[/tex]

Therefore, the height of the shrubs when they are sold (at t=7 years) can be found by substituting t=7 into the height function h(t).

[tex]h(7) = 0.65 * 7^2 + 2 * 7 + 17 = 52.85 cm[/tex]

So, the shrubs are 52.85 cm tall when they are sold.

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(a). The height of the shrub after t years is given by the function h(t) = (1.3/2)[tex]t^{2}[/tex] + 2t + 17.

(b). After 7 years, the shrubs reach a height of 62.85 cm.

Let's solve the given problem step-by-step:

(a)

The growth rate of the shrub is given by the differential equation dh/dt = 1.3t + 2. To find the height function h(t), we need to integrate this equation with respect to t.

1. We have:

dh/dt = 1.3t + 2

2. Integrating both sides with respect to t,

∫(dh) = ∫(1.3t + 2) dt

h(t) = (1.3/2)[tex]t^{2}[/tex]+ 2t + C

3. Where C is the constant of integration. To find C, we use the initial condition: h(0) = 17.

So:

17 = (1.3/2)[tex]0^{2}[/tex] + 2(0) + C

C = 17

4.Thus, the height function is:

h(t) = (1.3/2)[tex]t^{2}[/tex] + 2t + 17

(b)

To find the height after 7 years, we substitute t = 7 into the height function:

h(7) = (1.3/2)[tex]7^{2}[/tex] + 2(7) + 17

h(7) = (1.3/2)(49) + 14 + 17

h(7) = 31.85 + 14 + 17

h(7) = 62.85

Therefore, the shrubs are 62.85 cm tall when they are sold.

Answer:

1/12

Step-by-step explanation:

Probability is the result of the number of possible outcome divided by the number of total outcome.

Given that the box contains a red shirt, a blue shirt, a black trouser, and a red trouser, the number of total outcome is

= 1 + 1 + 1 + 1

= 4

The possibility of picking;

a black trouser is

= 1/4

then a red (without replacement, the total outcome drops to 3)

= 1/3

Hence, the probation choosing the black trouser and the red shirt without replacement

= 1/4 * 1/3

= 1/12

Answer:

I believe it's 188

Step-by-step explanation:

If there was a total of 564 and the cheeseburgers was three times the amount, you need to divide 564 by 3. 564 divided by 3 equals to 188. I am pretty sure this could be the correct answer!

Answer:

Step-by-step explanation:

The speed of light in a vacuum is 186,282 miles per second (299,792 kilometres per second), and in theory nothing can travel faster than light. In miles per hour, light speed is, well, a lot: about 670,616,629 mph. If you could travel at the speed of light, you could go around the Earth 7.5 times in one second.

186 ,282 x 5  = 931 , 410  for 5 seconds

Answer:

During the test for any significant difference in the means for the three assembly methods, there is significant evidence to reject the claim of equal population means

Step-by-step explanation:

Let;

SS be sum of square

MS be Mean Square

d.f be degree of freedom

trt be treatment

w be workers

Pls see attached files for detail step by step solutions as typing the explanation is not possible because of tables involved.

Using a one-way ANOVA test, it is determined that at least one of the assembly methods differs significantly from the others in terms of units assembled correctly by the randomly assigned workers.

The question involves the use of a one-way Analysis of Variance (ANOVA) to explore any significant difference in the means of the three proposed assembly methods. The given data are: SST (Total Sum of Squares) = 10800, SSTR (Sum of Squares between groups) = 4560. We can calculate SSE (Sum of Squares within groups) using the formula: SSE = SST - SSTR, which gives 10800 - 4560 = 6240.

The degrees of freedom (df) for between groups is k - 1 = 3 - 1 = 2 (where k is the number of groups). Similarly, the total degrees of freedom is n - 1 = 30 - 1 = 29 (where n is the total number of observations). The within groups degrees of freedom is df(total) - df(between) = 29 - 2 = 27.

Next, calculate Mean Square Between (MSB) = SSTR/df(between) = 4560/2 = 2280 and Mean Square Error (MSE) = SSE/df(within) = 6240/27 = 231.112. The F-statistic is given by F = MSB/MSE = 2280/231.112 = 9.869.

For a one-way ANOVA test at alpha = .05 with df(between) = 2 and df(within) = 27, checking an F distribution table, the critical value of F is approximately 3.354. Since our calculated F(9.869) > F critical(3.354), there is evidence to reject the null hypothesis. This indicates that at least one assembly method has a significantly different mean than the others.

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Answer:

B = 45

Step-by-step explanation:

B/3 - 6 = 9

B/3 = 15

B = 45

Tell me if I am wrong.

Can I get brainliest

Answer:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(58,21)[/tex]  

Where [tex]\mu=58[/tex] and [tex]\sigma=21[/tex]

The z score for this case is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

We know that the value of z = 1.78 and we want to estimate the value of X. If we solve for X from the z formula we got:

[tex] X= \mu +1.78 \sigma= 58 +1.78*21= 95.38[/tex]

So the corresponging weight for a z score of 1.78 is 95.38 grams

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(58,21)[/tex]  

Where [tex]\mu=58[/tex] and [tex]\sigma=21[/tex]

The z score for this case is given by this formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

We know that the value of z = 1.78 and we want to estimate the value of X. If we solve for X from the z formula we got:

[tex] X= \mu +1.78 \sigma= 58 +1.78*21= 95.38[/tex]

So the corresponging weight for a z score of 1.78 is 95.38 grams

A chicken egg with a z-score of 1.78 in a normal distribution model N(58, 21) will weigh approximately 95.38 grams.

The normal model N(58, 21) describes that weights of chicken eggs are normally distributed with a mean of 58 grams and a standard deviation of 21. Given a z-score of 1.78 for a randomly selected egg, we can calculate the egg's weight according to the Z-score formula, Z = (X - µ)/σ, where X is the value we want to find, µ is the mean, σ is the standard deviation, and Z is the z-score.

Let's apply the z-score value to the formula and solve for X:

Therefore, the weight of the egg with the z-score of 1.78 is approximately 95.38 grams.

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Answer:

50.265

Step-by-step explanation:

Sorry if its wrong

Answer:

87.74

Step-by-step explanation:

Answer:

a(1)=1

a(n)=a(n-1)+14

Step-by-step explanation:

The first term is 1, so a(1) = 1

Now, the second term, n=2 which means a(2)=a(1)+x

a(2)=1+x

also a(2)=15, then 1+x=15, and x=14

We can also check using a(3), a(4), and it fits.

The recursive formula of the arithmetic sequence is a(n) = a(n − 1) + 14.

There are two definitions for an arithmetic sequence. It is a "series where the differences between every two succeeding terms are the same" or "each term in an arithmetic sequence is formed by adding a fixed number (positive, negative, or zero) to its preceding term."

Given arithmetic sequence 1, 15, 29, 43, ....

the first term of the sequence is 1

a(1) = 1

and formula a(n) = a(n − 1) + x

the second term is 15

a(2) = 15 and n = 2

substitute in formula

a(n) = a(n − 1) + x

a(2) = a(2 - 1) + x

15 = a(1) + x

15 = 1 + x

x = 14

so formula is  a(n) = a(n − 1) + 14

check for n = 3

a(3) = a(3 - 1) + 14

a(3) = a(2) + 14

a(3) = 15 + 14 = 26

Hence the formula is  a(n) = a(n − 1) + 14

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Step-by-step explanation:

Area, A = Length,x × Breadth,y

A=xy

When x and y are independent, E(xy) = E(x)E(y)

As x and y have the same distribution, U[L-A,L+A], they have the same mean.

We could argue by symmetry that E(x) = L and E(y) + L, also.

We can also reason this from the fact that, if X ~ U[L-A, L+A], f(x) = 1/(2A) from L-A to L+A

Therefore

[tex]E(x)= \int\limits {f(x)} \, dx \\\\=\int\limits^{(L+A)}_{(L-A)} {\frac{1}{2A}x } \, dx[/tex]

[tex]=\int\limits^{(L+A)}_{(L-A)} {\frac{1}{2A}x } \, dx \\\\=[\frac{1}{4A}x^2 ]\limits^{(L+A)}_{(L-A)}[/tex]

=1/(4A)(L+A)2 - 1/(4A)(L-A)2

= 1/(4A)(L2 + 2AL + A2) - 1/(4A)(L2 - 2AL + A2)

=1/(4A)(2AL+2AL)

= 1/(4A)(4AL)

= L

Thus, E(xy) = E(x)E(y) = L×L = L²

Answer:

-0.36666666666...

Step-by-step explanation:

not sure how to explain, that's a repeating number.

Answer:

10

Step-by-step explanation:

hope it helps

Answer:

5000m

12/3 = 4yd

2kg

8x16 = 128oz

9000ml

20qt

120+23 =143mins

Answer:

5000m

4yd

2kg

128oz

9000ml

20qt

143min

Answer:

The standard error( X1 − X2 ) = 0.547

Step-by-step explanation:

Step:-(1)

Given a procurement specialist has purchased 23 resistors

Given normally and independently distributed with mean 120 ohms and standard deviation 1.7

mean of the Population of the vendor 1 is μ₁ =  120 ohms

Standard deviation of the Population the vendor 1 is σ₁ = 1.7 ohms

similarly represent the vendor 2 observed resistances, which are assumed to be normally and independently distributed with mean 125 ohms and standard deviation of 2.0

mean of the Population of the vendor 2 is μ₂ =  120 ohms

Standard deviation of the Population the vendor 2 is σ₂ = 1.7 ohms

The standard error of the  difference of two means

 Se( X1 − X2) = [tex]\sqrt{\frac{σ^2_{1} }{n_{1} } +\frac{σ^2_{2} }{n_{1} } }[/tex]

 Here σ₁ = 1.7 ohms and σ₂ = 2 ohms and n₁=n₂ =n = 23 resistors

se(X1 − X2) = [tex]\sqrt{\frac{1.7^2}{23 } +\frac{2^2 }{23} }[/tex]

se(X1 − X2) = √0.2995

                   = 0.547

Conclusion:-

The standard error of X1 − X2 = 0.547

Final answer:

Explains the standard error of sample means for two vendors and the distribution of sample means.

Explanation:

The standard error of the sample mean in this scenario, rounded to two decimal places, is calculated as follows:

For vendor 1: standard error = 1.7 / sqrt(23)

For vendor 2: standard error = 2.0 / sqrt(30)

The distribution of the sample mean X: Since the sample means are normally distributed, the sampling distribution of X is approximately normal.

The standard error of X1 − X2: The standard error of the difference between sample means X1 and X2 is calculated using the formula for the standard error of the difference between two independent sample means.

I suppose x should be π.

Recall the double angle identity for cosine:

[tex]\cos^2\dfrac x2=\dfrac{1+\cos x}2[/tex]

Then remember for [tex]0<x<\frac\pi2[/tex], we have [tex]\cos x>0[/tex].

Let [tex]x=\frac{7\pi}6[/tex]. Plugging this into the equation above gives

[tex]\cos^2\dfrac{7\pi}{12}=\dfrac{1+\cos\frac{7\pi}6}2[/tex]

Take the square root of both sides; this introduces two possible values, but we know [tex]\cos\frac{7\pi}{12}[/tex] should be positive, so

[tex]\cos\dfrac{7\pi}{12}=\sqrt{\dfrac{1+\cos\frac{7\pi}6}2}=\dfrac{\sqrt{2-\sqrt3}}2[/tex]

Answer:

Step-by-step explanation:

Reduction to normal from using lambda-reduction:

The given lambda - calculus terms is, (λf. λx. f (f x)) (λy. Y * 3) 2

For the term, (λy. Y * 3) 2, we can substitute the value to the function.

Therefore, applying beta- reduction on "(λy. Y * 3) 2" will return 2*3= 6

So the term becomes,(λf. λx. f (f x)) 6

The first term, (λf. λx. f (f x)) takes a function and an argument, and substitute the argument in the function.

Here it is given that it is possible to substitute the resulting multiplication in the result.

Therefore by applying next level beta - reduction, the term becomes f(f(f(6)) (f x)) which is in normal form.

Answer:

Cube and cuboid

Both have 6 faces

Answer:

Cube and cuboid

Both have 6 faces

The area of a circle can be calculated by formula πr^2...

radius(r)=4inches

therefore,

Area(A)=3.14×4

=12.56

Final answer:

To calculate the area of a circle with an 8-inch diameter using 3.14 for π, the formula A = πr² gives us 50.24 square inches, which rounds to 50.2 square inches to the nearest tenth.

Explanation:

The student is asking for the area of a circle with a diameter of 8 inches using 3.14 for π (Pi). The area of a circle is given by the formula A = πr², where 'r' is the radius of the circle. Since the diameter is twice the radius, we have that the radius 'r' is 4 inches (half of 8 inches). Plugging this into the formula we get A = 3.14 × 4².

To find the area to the nearest tenth, we perform the calculation: A = 3.14 × 16 = 50.24. Therefore, the area of the circle is 50.2 square inches when rounded to the nearest tenth.

========================================================

Explanation:

One way to go about this is to list out all the perfect cubes. A perfect cube is the result of taking any whole number and multiplying it by itself 3 times.

1 cubed = 1^3 = 1*1*1 = 1

2 cubed = 2^3 = 2*2*2 = 8

3 cubed = 3^3 = 3*3*3 = 27

4 cubed = 4^3 = 4*4*4 = 64

and so on. We stop once we reach 1721, or if we go over. Ignore any values larger than 1721. You'll find that 11^3 = 1331 and 12^3 = 1728. So we stop here and exclude 1728 as that is larger than 1721.

A quick way to see where we should stop is to apply the cube root to 1721 and we get

[tex]\sqrt[3]{1721} = 1721^{1/3} \approx 11.98377[/tex]

The approximate result of 11.98377 tells us that 1721 is between the perfect cubes of 11^3 = 1331 and 12^3 = 1728

------------------

So effectively, we have 11 perfect cubes in the set {0, 1, 2, 3, ..., 1721} and this is out of 1722 numbers in that same set. Note how I added 1 onto 1721 to get 1722. I'm adding an extra number because of the 0. If 0 wasn't part of the set, then we would have 1721 values total inside.

In summary: There are 11 values we want (11 perfect cubes) out of 1722 values total.

Divide 11 over 1722 to get

11/1722 = 0.00638792102207

which rounds to 0.006388

The probability of randomly selecting a perfect cube from the set {0,1,2,...,1721} is calculated by finding the ratio of the number of favorable outcomes (perfect cubes in this range) to the total number of outcomes. We have 12 perfect cubes and 1722 total numbers, giving us a probability of approximately 0.006964.

The subject of this question is probability, a concept in mathematics. A perfect cube is a number that can be expressed as the cube of an integer.For instance, the numbers 1 (13), 8 (23), 27 (33), and so forth, are perfect cubes.

In this case, we need to find how many perfect cubes exist between 0 and 1721. The cube of 12 is 1728, which is greater than 1721 so, we can conclude that the largest whole number whose cube is less than 1721 is 11. This means there are 12 (perfect cubes) numbers from 0 to 1721 (0 included).

The total amount of numbers in this range is 1722 (from 0 to 1721 inclusive). Therefore, the probability of randomly selecting a perfect cube from this range is the ratio of the number of favorable outcomes (perfect cubes) to the total number of outcomes:

P(perfect cube) = Number of perfect cubes /Total numbers

P(perfect cube) = 12/1722 ≈ 0.006964.

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Step-by-step explanation:

R'(x) = 5x³ − 60x²

A) R"(x) = 15x² − 120x

0 = 15x² − 120x

0 = 15x (x − 8)

x = 0 or 8

B) R"(-1) = 135

R"(1) = -105

R"(x) changes signs around x = 0, so x = 0 is an inflection point.

R"(7) = -105

R"(9) = 135

R"(x) changes signs around x = 8, so x = 8 is an inflection point.

Answer:

Step-by-step explanation:

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 35

For the alternative hypothesis,

µ > 35

It is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 16,

Degrees of freedom, df = n - 1 = 16 - 1 = 15

t = (x - µ)/(s/√n)

Where

x = sample mean = 42.7

µ = population mean = 32

s = samples standard deviation = 6

t = (42.7 - 32)/(6/√16) = 7.13

We would determine the p value using the t test calculator. It becomes

p = 0.00001

Since alpha, 0.01 > than the p value, 0.00001, then we would reject the null hypothesis. Therefore, At a 1% level of significance, there is sufficient data to conclude that children with a history of child care show significantly more behavioral problems than the average kindergarten child

b) Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

the information given, the from population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution,

Since confidence level = 90% = 0.95, α = 1 - CL = 1 – 0.90 = 0.1

α/2 = 0.1/2 = 0.05

the area to the left of z0.05 is 0.05 and the area to the right of z0.05 is 1 - 0.05 = 0.95

Looking at the t distribution table for t.95 and df = 15

z = 1.753

Margin of error = 1.753 × 6/√16

= 2.63

Confidence interval = 35 ± 2.63

Answer:

A. Financial data tapes that contain data compiled from the New York Stock

Step-by-step explanation:

Primary data is data that is collected by a researcher from first-hand sources. Data published by the New York Stock Exchange is primary data, which inherently makes the data tapes secondary data, because they contain the data from a source other than original source.

The total value of the German shareholding is 202500 million dollars.

Given that, the total number of shares is 45 million.

We need to find the total value of the German shareholding.

Germany holds 15% of the shares, which is 15% of 45 million

=15/100×45000000=67,50,000

Number of shares Germany holds=67,50,000

Now, the total value of the German shares=67,50,000×30=$20,25,00,000

Therefore, the total value of the German shareholding is 202500 million dollars.

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Final answer:

The total value of the German shareholding in Tazon Insurance is $202.5 million as of April in Year 1.

Explanation:

The student is asking for the calculation of the total value of German shareholdings in Tazon Insurance when each share is valued at $30. To find this, we need to calculate the number of shares held by Germany and then multiply by the share price.

Therefore, the total value of the German shareholding in Tazon Insurance is $202.5 million as of April in Year 1.

Answer:

[tex] P(\hat p>0.14)[/tex]

And using the z score given by:

[tex] z = \frac{\hat p -\mu_p}{\sigma_p}[/tex]

Where:

[tex]\mu_{\hat p} = 0.12[/tex]

[tex]\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149[/tex]

If we find the z score for [tex]\hat p =0.14[/tex] we got:

[tex]z = \frac{0.14-0.12}{0.0149}= 1.340[/tex]

So we want to find this probability:

[tex] P(z>1.340)[/tex]

And using the complement rule and the normal standard distribution and excel we got:

[tex] P(Z>1.340) = 1-P(Z<1.340) = 1-0.9099= 0.0901[/tex]

Step-by-step explanation:

For this case we have the proportion of interest given [tex] p =0.12[/tex]. And we have a sample size selected n = 474

The distribution of [tex]\hat p[/tex] is given by:

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

We want to find this probability:

[tex] P(\hat p>0.14)[/tex]

And using the z score given by:

[tex] z = \frac{\hat p -\mu_p}{\sigma_p}[/tex]

Where:

[tex]\mu_{\hat p} = 0.12[/tex]

[tex]\sigma_{\hat p}= \sqrt{\frac{0.12*(1-0.12)}{474}}= 0.0149[/tex]

If we find the z score for [tex]\hat p =0.14[/tex] we got:

[tex]z = \frac{0.14-0.12}{0.0149}= 1.340[/tex]

So we want to find this probability:

[tex] P(z>1.340)[/tex]

And using the complement rule and the normal standard distribution and excel we got:

[tex] P(Z>1.340) = 1-P(Z<1.340) = 1-0.9099= 0.0901[/tex]