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erma4kov [3.2K]

3 years ago

5

Explain why the prMarco Invests $760 into a savings account. The account pays 4% simple interest. If Marco makes no deposits or

withdrawals to his account in 5 years. How much will it be in that account?

1 answer:

Vladimir79 [104]3 years ago

8 0

Answer:

Step-by-step explanation:

start with 760 dollars. find 4% of that amount then add that % plus the starting amount 5 diffrent times.

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This should be easy

Ivanshal [37]

Answer:

69

Step-by-step explanation:

it's the holy number jesus christ can confirm

8 0

2 years ago

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Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology ex

Nataly [62]

Answer:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

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".

$p_A$ represent the real population proportion female for Biology

$\hat p_A =\frac{84199}{144796}=0.582$ represent the estimated proportion female for biology

$n_A=144796$ is the sample size for A

$p_B$ represent the real population proportion female for calculus AB

$\hat p_B =\frac{102598}{211693}=0.485$ represent the estimated proportion female for Calculus AB

$n_B=211693$ is the sample size required for B

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$

For the 90% confidence interval the value of $\alpha=1-0.90=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$(0.582-0.485) - 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.0942$

$(0.582-0.485) + 1.64 \sqrt{\frac{0.582(1-0.582)}{144796} +\frac{0.485(1-0.485)}{211693}}=0.09978$

And the 90% confidence interval would be given (0.0942;0.09978).

We are confident at 90% that the difference between the two proportions is between $0.0942 \leq p_A -p_B \leq 0.09978$

5 0

3 years ago

If you have 23500 apples and you had 42600 oranges how many do you have in all?

vekshin1

Answer:

66,100 in all.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce

34kurt

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(boxes weigh more than 22.9 ounces)

P(x > 22.9)

$P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)$

$= 1 - P(z \leq -2)$

Calculation the value from standard normal z table, we have,

$P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%$

97.7% of of the boxes weigh more than 22.9 ounces.

b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

$P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)$

Calculation the value from standard normal z table, we have,

$P(x < 23.7) =0.159= 15.9\%$

15.9% of of the boxes weigh less than 23.7 ounces.

7 0

3 years ago

Can someone tell me if I’m right? I put the 80 under 8

cupoosta [38]

Answer:

80

4

Step-by-step explanation:

You are correct in placing 80 in the box in the table. If you examine the table and the pattern, you'll see that as the number of marbles increases by 10, the weight of the bags increases by 40.

We can determine the rule by dividing 40 by 10: 40 ÷ 10 = 4.

So, the rule is actually to divide the weight of the bag by 4 to find the number of marbles.

Note that this is likely because each marble is 4 grams.

6 0

3 years ago

Read 2 more answers