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aleksklad [387]
3 years ago
6

HELP QUICK PLEASE, NEED ANSWERS TO ALL 3

Mathematics
1 answer:
alukav5142 [94]3 years ago
4 0

Answer:

 7.5 divided by 15% = 50

$12,500 divided by 50 = $250

f(t)=7.5 divided by 15% =50 divided by $12,500 = $250

4~

good luck <3

! ( make sure where I said "divided by" you put division signs!!) !

Step-by-step explanation:

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Mr. Jamison deposited $100 into a new savings account on January 1. On the first day of each month thereafter, he deposited thre
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$36 400

Step-by-step explanation:

Step 1

The first step is to figure out how much money is saved at the end of each month for the period from January 1 to June 15. The amount deposited at the end of each month is obtained by multiplying the amount from the previous month by 3.

The amount deposited in January is  \$100.

The amount deposited in February is 1\$00\times 3= \$300.

The amount deposited in March is  \$300\times 3= \$900.

The amount deposited in April is  \$900\times 3= \$2\700.

The amount deposited in May is  \$2\,700\times 3= \$8\,100.

The amount deposited in June is  \$8\,100\times 3= \$24\,300.


Step 2

The next step is to add up all the money that was deposited into the account. This calculation is shown below,

\$100+\$300+900+\$\$2\,700+\$8\,100+\$24\,000=\$36\,400


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Answer:

-12 + 3 - (10x - 15x + 96)

Step-by-step explanation:

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Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic t.
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Using the t-distribution, it is found that the p-value of the test is 0.007.

At the null hypothesis, it is <u>tested if the mean lifetime is not greater than 220,000 miles</u>, that is:

H_0: \mu \leq 220000

At the alternative hypothesis, it is <u>tested if the mean lifetime is greater than 220,000 miles</u>, that is:

H_1: \mu > 220000.

We have the <u>standard deviation for the sample</u>, thus, the t-distribution is used. The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

For this problem:

\overline{x} = 226450, \mu = 220000, s = 11500, n = 23

Then, the value of the test statistic is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{226450 - 220000}{\frac{11500}{\sqrt{23}}}

t = 2.69

We have a right-tailed test(test if the mean is greater than a value), with <u>t = 2.69</u> and 23 - 1 = <u>22 df.</u>

Using a t-distribution calculator, the p-value of the test is of 0.007.

A similar problem is given at brainly.com/question/13873630

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