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Ilya [14]
3 years ago
8

How much money would need to be deposited into an account earning 5.75% interest compounded annually in order for the accumulate

d value at the end of 25 years to be $85,000?
Mathematics
2 answers:
Leona [35]3 years ago
6 0
I will assume you are using compound interest. 

<span>let the amount invested be x </span>

<span>x(1.0575)^25 = 85000 </span>
<span>x = 85000/1.0575^25 = $21,009.20</span>
Semmy [17]3 years ago
3 0

Answer:

The money that would need to be deposit into the account is $21009.44

Step-by-step explanation:

Given:  Interest = 5.75% compounded annually

Amount = $ 85,000

Times period = 25 years

We have to calculate the money that would need to be deposit into the account.

We know the formula for compound interest

A=P(1+\frac{r}{100})^n

Where, A is amount

P is principal amount

n is time

r = rate of interest

Thus, Substitute, we get,

85000=P(1+\frac{5.75}{100})^25

Solving for P,

\left(1+\frac{5.75}{100}\right)^{25}=4.0458(approx)

Divide both side by 4.0458, we get,

P=\frac{85000}{4.0458}=21009.44

Thus, the money that would need to be deposit into the account is $21009.44

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Sequence 1,5,9,13,...

A(0) = 1 +4x0=1

A(1) =1 + 4x1 = 5

A(2) = 1+ 4x2= 9

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4 0
3 years ago
For each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts. A large shipmen
Stella [2.4K]

Answer:

(a) The p-value of the test statistic should be 0.33.

(b) No, there is not sufficient evidence to reject the entire shipment.

Step-by-step explanation:

We are given that for each shipment of parts a manufacturer wants to accept only those shipments with at most 10% defective parts.

A quality control manager randomly selects 50 of the parts from the shipment and finds that 6 parts are defective.

Let p = <u><em>proportion of defective parts among the current shipment.</em></u>

So, Null Hypothesis, H_0 : p \leq 10%      {means that the defective parts in the shipment is at most 10%}

Alternate Hypothesis, H_A : p > 10%      {means that the defective parts in the shipment is greater 10%}

The test statistics that would be used here <u>One-sample z proportion</u> <u>statistics</u>;

                      T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of parts that are defective among the current shipment = \frac{6}{50} = 12%

           n = sample of parts from the shipment = 50

So, <u><em>test statistics</em></u>  =  \frac{0.12-0.10}{\sqrt{\frac{0.12(1-0.12)}{50} } }

                               =  0.44

The value of z test statistics is 0.44.

(a) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P(Z > 0.44) = 1 - P(Z \leq 0.44)

                        = 1 - 0.67003 = 0.3299 ≈ 0.33

(b) Since, the P-value of test statistics is more than the level of significance as 0.33 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the defective parts in the shipment is at most 10%.

4 0
3 years ago
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