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Rasek [7]
2 years ago
12

What is 2 hundreds + 15 tens + 6 ones

Mathematics
2 answers:
qwelly [4]2 years ago
5 0
Two one hundred is 200.
15 tens is 150.
Six ones is the same.
200 + 150 + 6 = 456
valentinak56 [21]2 years ago
3 0
=356// 200+150 (15x10)+6 //
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lorasvet [3.4K]
4065323 × 10 to the senventh power?
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Javier deposited $250 into a savings account with an interest rate of 1.6%. He made no deposits or withdrawals for 9 months. If
11Alexandr11 [23.1K]

Answer:

A = 250(1 + 0.016)^0.75.

Step-by-step explanation:

9 months = 0.75 years

So A = 250(1 + 0.016)^0.75.

4 0
2 years ago
According to the U.S. Census Bureau (2016), a substantial pay gap still exists. The average American man, with an advanced colle
Leokris [45]

Answer:

The difference in earnings, over a 30-year career, for men vs women, is $1,200,150

Step-by-step explanation:

Per year.

The average man earns $90,761.

The average woman earns $50,756

So, per year, the difference is:

90,761 - 50,756 = 40,005

Over 30 years:

30*40,005 = 1,200,150

The difference in earnings, over a 30-year career, for men vs women, is $1,200,150

0 0
2 years ago
Z to the power of 2 =1
stiv31 [10]

That would be written as:

z^{2} = 1


And the step-by-step equation would be:

z^{2} = 1 → we take the 2 to the right side of the equation and do the IO*

z = \sqrt{1} →we solve for \sqrt{1}

z = 1 → final answer


*Ps - IO is not an existing term and stands for inverse operation. In this case, because when we take to 2 to the right side of the equation (the 2 is a power) it'll have to turn into a square root (because exponents and roots are inverse operations)



Hope it helped,



BioTeacher101

3 0
2 years ago
Suppose babies born in a large hospital have a mean weight of 4095 grams, and a standard deviation of 569 grams. If 130 babies a
Butoxors [25]

Answer:

P =0.3998

Step-by-step explanation:

Let {\displaystyle {\overline{x}}} be the average of the sample, and the population mean will be \mu

We know that:

\mu = 4095 gr

Let \sigma be the standard deviation and n the sample size, then we know that the standard error of the sample is:

E=\frac{\sigma}{\sqrt{n}}

Where

\sigma=569

n=130

In this case we are looking for:

P(|{\displaystyle{\overline{x}}}- \mu|>42)

This is:

{\displaystyle{\overline{x}}}- \mu>42 or {\displaystyle{\overline{x}}}- \mu

P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu

Now we get the z score

Z=\frac{{\displaystyle{\overline{x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}

P=P(z>\frac{42}{\frac{569}{\sqrt{130}}}) + P(z

P=P(z>0.8416) + P(z

Looking at the tables for the standard nominal distribution we get

P =0.1999+0.1999

P =0.3998

6 0
2 years ago
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