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3 years ago

10

What is the answer to -4.3g=25.8

1 answer:

8090 [49]3 years ago

7 0

G=-6
You just divide both sides by -4.3

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A study involved measuring the average IQ of Americans. What does the Central Limit Theorem say about the study? As long as the

Alexxx [7]

Answer:

As long as the sample size n is large enough: The average IQ of Americans in the sample will be normally distributed.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

So the correct answer is:

As long as the sample size n is large enough: The average IQ of Americans in the sample will be normally distributed.

5 0

3 years ago

What is the formula of circle

gogolik [260]

Answer:

Area = pi r squared

Step-by-step explanation:

If you need to, sometimes your teacher will ask you to leave you answers in terms of pi, or keep pi as 3.

3 0

2 years ago

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Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

I need help with this, I don’t understand how to do this

DanielleElmas [232]

Answer:

76 because the angles must add up to 360

5 0

3 years ago

Read 2 more answers

Someone please help fast

Bad White [126]

Answer:

16/3 because you put a 1 under 8 since it's not a fraction hen multiple across

Step-by-step explanation:

pls give brainly

4 0

3 years ago

Read 2 more answers