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3 years ago

Suppose you buy a car with a value of 12,500.Each year your car depreciates by 3.6%.How much will your car be worth in 5 years

1 answer:

6 0

So the multiplier for decreasing by 3.6% is 0.964 (1 - 0.036)
Times the value by the multiplier to the power of 5, since that is the number of years you want.

12,500 x 0.964^5 = 10,406.27 (2dp)

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Four students are determining the probability of flipping a coin and it landing head's up. Each flips a coin the number of times

Oxana [17]

Answer:

Collin

Step-by-step explanation:

He did the most attempts.

The probability of flipping a coin is 1/2

7 0

3 years ago

Read 3 more answers

Refer to the following scenario:You want to see if there is a difference between the exercise habits of Science majors and Math

bekas [8.4K]

Answer:

1. H0: P1 = P2

2. Ha: P1 ≠ P2

3. pooled proportion p = 0.542

4. P-value = 0.0171

5. The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

6. The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

Step-by-step explanation:

We should perform a hypothesis test on the difference of proportions.

As we want to test if there is significant difference, the hypothesis are:

Null hypothesis: there is no significant difference between the proportions (p1-p2 = 0).

Alternative hypothesis: there is significant difference between the proportions (p1-p2 ≠ 0).

The sample 1 (science), of size n1=135 has a proportion of p1=0.607.

$p_1=X_1/n_1=82/135=0.607$

The sample 2 (math), of size n2=92 has a proportion of p2=0.446.

$p_2=X_2/n_2=41/92=0.446$

The difference between proportions is (p1-p2)=0.162.

$p_d=p_1-p_2=0.607-0.446=0.162$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{82+41}{135+92}=\dfrac{123}{227}=0.542$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.542*0.458}{135}+\dfrac{0.542*0.458}{92}}\\\\\\s_{p1-p2}=\sqrt{0.001839+0.002698}=\sqrt{0.004537}=0.067$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.162-0}{0.067}=\dfrac{0.162}{0.067}=2.4014$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$\text{P-value}=2\cdot P(z>2.4014)=0.0171$

As the P-value (0.0171) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a signficance level of 0.01, there is not enough evidence to support the claim that there is significant difference between the exercise habits of Science majors and Math majors .

We want to calculate the bounds of a 99% confidence interval of the difference between proportions.

For a 99% CI, the critical value for z is z=2.576.

The margin of error is:

$MOE=z \cdot s_{p1-p2}=2.576\cdot 0.067=0.1735$

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

$LL=(p_1-p_2)-z\cdot s_{p1-p2} = 0.162-0.1735=-0.012\\\\UL=(p_1-p_2)+z\cdot s_{p1-p2}= 0.162+0.1735=0.335$

The 99% confidence interval for the difference between proportions is (-0.012, 0.335).

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3 years ago

1000 households were surveyed. 275 households own a desktop computer, 455 households own a DVD player, 405 households own two ca

icang [17]

Answer:

Step-by-step explanation:

From the given information,

Suppose

X represents the Desktop computer

Y represents the DVD Player

Z represents the Two Cars

Given that:

n(X)=275

n(Y)=455

n(Z)=405

n(XUY)=145

n(YUZ)=195

n(XUZ)=110

n((XUYUZ))=265

n(X ∩ Y ∩ Z) = 1000-265

n(X ∩ Y ∩ Z) = 735

n(X ∪ Y) = n(X)+n(Y)−n(X ∩ Y)

145 = 275+455 - n(X ∩ Y)

n(X ∩ Y) = 585

n(Y ∪ Z) = n(Y) + n(Z) − n(Y ∩ Z)

195 = 455+405-n(Y ∩ Z)

n(Y ∩ Z) = 665

n(X ∪ Z) = n(X) + n(Z) − n(X ∩ Z)

110 = 275+405-n(X ∩ Z)

n(X ∩ Z) = 570

a. n(X ∪ Y ∪ Z) = n(X) + n(Y) + n(Z) − n(X ∩ Y) − n(Y ∩ Z) − n(X ∩ Z) + n(X ∩ Y ∩ Z)

n(X ∪ Y ∪ Z) = 275+455+405-585-665-570+735

n(X ∪ Y ∪ Z) = 50

c. n(X ∪ Y ∪ C') = n(X ∪ Y)-n(X ∪ Y ∪ Z)

n(X ∪ Y ∪ C') = 145-50

n(X ∪ Y ∪ C') = 95

6 0

3 years ago

The graph shows a distribution of data.

ss7ja [257]

Answer:

.0625 or A

Step-by-step explanation:

You have to work out the mean. (The simple average of those numbers) Then for each number subtract the Mean and square the result. Then work out the average of those squared differences.

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3 years ago

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As the student council treasurer, you prepare the budget for your class rafting trip. Each large raft costs $100 to rent and eac

Art [367]

Answer- You can get 10 Small rafts

12 x 100+ 1200

1600-1200=400

400/40 =10 Rafts

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3 years ago