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erma4kov [3.2K]
3 years ago
11

marco deposited $500 in a savings account earning 4% simple intrest over 5 years what was the total amount of intrest earned at

the end of 5 years a. 40, b. 50, c. 100, d.200
Mathematics
1 answer:
tatuchka [14]3 years ago
7 0

The formula in computing the simple interest of a savings account is:

I = Prt

Where I = interest earned

             P = principal amount deposited

             r = rate

             t = time in years

You are asked to find the total amount of interest earned at the end of 5 years. Substituting the amounts to the formula:

I = $500 (4%) (5)

I = $100

Therefore, the total amount of interest earned at the end of 5 years is $100.

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Part 2

7-8\div4(3^2-1)=7-8\div4(9-1) \\  \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\  \\ =6  \frac{1}{2} = \frac{13}{2}



Part 3

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Part 4

(5^2-9\cdot3)^2-11=(25-27)^2-11 \\  \\ =(-2)^2-11=4-11=-7



Part 5

[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\  \\ =(1-5)\cdot4=-4\cdot4=-16



Part 6

12-11\cdot2+16\div8=12-22+2 \\  \\ =-8



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Part 12

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A laptop producing company also produces laptop batteries, and claims that the batteries
gregori [183]

Answer:

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours. (P-value = 0).

The null and alternative hypothesis are:

H_0: \mu=4\\\\H_a:\mu< 4

Step-by-step explanation:

<em>The question is incomplete: To test this claim a sample or population standard deviation is needed.</em>

<em>We will estimate that the sample standard deviation is 2 hours, and use a t-test to test that claim.</em>

<em> NOTE (after solving): The difference between the sample mean and the mean of the null hypothesis is big enough to reject the null hypothesis, even when we have a sample standard deviation of 3.5 hours, which can be considered bigger than the maximum standard deviation for the sample.</em>

This is a hypothesis test for the population mean.

The claim is that the batteries power the laptops for significantly less than 4 hours.

Then, the null and alternative hypothesis are:

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The significance level is 0.05.

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The sample mean is M=3.5.

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This test is a left-tailed test, with 499 degrees of freedom and t=-5.5902, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the batteries power the laptops for significantly less than 4 hours.

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