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3 years ago

Leta has $45 in her account in May how much money does she have in her account in august

Mathematics

2 answers:

Zepler [3.9K]3 years ago

6 0

I think it would be $22

inessss [21]3 years ago

5 0

I don’t know but ask google

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Student Debt – Vermont: The average student loan debt of a U.S. college student at the end of 4 years of college is estimated to

mezya [45]

Answer:

The confidence interval is (24116.3878,24883.6122).

Step-by-step explanation:

We are given the following information in the question:

Population mean, $\mu$ = $23,500

Sample mean, $\bar{x}$ = $24,500

Sample standard deviation,s = $2,800

Sample size, n = 146

Confidence interval:

$\bar{x} \pm t_{critical}\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at}~\alpha_{0.05} = \pm 1.655436$

$24500 \pm 1.65543(\frac{2800}{\sqrt{146}} ) = 24500 \pm 383.6122 = (24116.3878,24883.6122)$

The confidence interval is (24116.3878,24883.6122).

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3 years ago

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garik1379 [7]

I believe that the answer is a

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3 years ago

How to solve (2.31*10^-6)+(5.87*10^-4)

S_A_V [24]

We want to calculate the following

$2.31\cdot10^{-6}+5.87\cdot10^{-4}$

Using the properties of exponents, we have that

$10^{-6}=10^{-4}\cdot10^{-2}$

So we have

$2.31\cdot10^{-6}+5.87\cdot10^{-4}=2.31\cdot10^{-2}\cdot10^{-4}+5.87\cdot10^{-4}$

So, if factor 10^-4 on the right side, we have

$10^{-4}(2.31\cdot10^{-2}+5.87)$

Note that

$2.31\cdot10^{-2}=0.0231$

Then,

$5.87+2.31\cdot10^{-2}=5.87+0.0231=5.8931^{}$

So we have that

$2.31\cdot10^{-6}+5.87\cdot10^{-4}=5.8931\cdot10^{-4}$

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1 year ago

The distance between -4,1 and -2,2

Vlad1618 [11]

Answer:

✓32

4✓2

5.66

Step-by-step explanation:

That's the answer. Hope it helps

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3 years ago

Whats the easiest way to do long division

Anastaziya [24]

Answer:

Set up the equation. On a piece of paper, write the number being divided on the right, under the division symbol, and the number doing the division to the left on the outside. ...

Divide the first digit. ...

Divide the first two digits. ...

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hope this helps :)

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4 years ago