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4 years ago

Help me out this question i dont understand it

Mathematics

2 answers:

SVEN [57.7K]4 years ago

4 0

Answer

I think it's a. 12%

Step-by-step explanation:

vitfil [10]4 years ago

3 0

I think it’s 12% or 28%

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Add the two expressions.<br><br> 3y + 5 and y + 24<br><br> Enter your answer in the box.

grigory [225]

Answer:

4y+29

Step-by-step explanation:

Collect like terms

3y + y = 4y

5 + 24 = 29

4y+29

4 0

3 years ago

A system of equations is shown. y = 2x - 7 y = -x + 5 What is the solution to the system of equations? Please Show your work

elixir [45]

Are u sure u didnt write this wrong? maybe it’s x=2y-7? this doesn’t make sense :(

4 0

3 years ago

A taxi travels 25 kilometers east of an airport. then, it travels from that point to a point that is 40 kilometers west of an ai

sashaice [31]

25km east+25km return+40km west+40km return = 130km

4 0

4 years ago

Read 2 more answers

I need help, i get like 4712, but its not an answer

ella [17]

So you will need to find the height, and we can do so by using the pythagorean theorem, which is $leg^2+leg^2=hypotenuse^2$ . Using this formula, we can make the equation $15^2+h^2=25^2$

Firstly, solve the exponents: $225+h^2=625$

Next, subtract 225 on both sides: $h^2=400$

Lastly, square root both sides, and your answer is h = 20.

Now that we got the height, we can solve for the volume, which is $V=\pi r^2\frac{h}{3}$ . Using this formula, we can form our equation to be $V=\pi 15^2\frac{20}{3}$

Firstly, solve the exponent: $V=\pi*225*\frac{20}{3}$

Next, multiply 225 and 20/3, and your answer will be V = 1500π, or the third option.

7 0

3 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

3 years ago