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3 years ago

Find the value of please help me

Mathematics

1 answer:

saveliy_v [14]3 years ago

6 0

Answer:

10 sqrt(10)

Step-by-step explanation:

sqrt(1000)

sqrt(100*10)

we know that sqrt(a*b) = sqrt(a) sqrt(b)

sqrt(100) * sqrt(10)

10 sqrt(10)

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Can you solve (4x-3)/(x+2)+(15)/(x-3)

Murljashka [212]

Please see the pic, I'd solved in it.

8 0

3 years ago

Read 2 more answers

Work out the volume of this prism.

a_sh-v [17]

Answer:

840

Step-by-step explanation:

Lets split the shape in 2 shapes which gives us 2 rectangular prisms

1st shape is the one stcking upward: 12*5*6=360

2nd shape: (we must subtract 5 from 15 because we used that for the other shape and hat wouldn't give us the correct answer) 10*8*6=480

480+360=

840

8 0

3 years ago

There are two square roots of 25. Which best explains why there is only one answer for the radius of the button?

Brrunno [24]

5 times5 is the one 5 times itself is 25 hents 25

4 0

3 years ago

A baby weighed 7.25 at birth. at the end of 8 month,the baby weighed 2 1/2 times its birth weight. how many pounds did the baby

Artyom0805 [142]

2 1/2 is in mixed fraction. So let's first make it into improper fraction.

(2*2 + 1) / 2 = 5/2

5/2 of 7.25 = (5*7.25) / 2 = 18.125

in short, 18.125 is the answer

4 0

3 years ago

A population has a standard deviation of 5.5. What is the standard error of the sampling distribution if the sample size is 81?

VladimirAG [237]

Answer:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

Step-by-step explanation:

For this case we know the population deviation given by:

$\sigma = 5.5$

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution $\bar X$ and for this case we know that the distribution is given by:

$\bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})$

And the standard error would be:

$\sigma_{\bar x}= \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$\sigma_{\bar x}=\frac{5.5}{\sqrt{81}}= 0.611$

4 0

3 years ago