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2 years ago

13

For THE LAST TIME CAN SOMEONE PLEASE I BEG YOU AND HELP ME BEFORE I FAINT BECAUSE I HAVE BEEN WORKING ON THIS SINCE THE MORNING

AND I REALLY NEED HELP PLEASE HELP MEEEEE THANK YOU

1 answer:

Lynna [10]2 years ago

3 0

I gotchu my guy let me know what you need help with I’d gladly revive those free points

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I JUST NEED #13 please and thnx

emmainna [20.7K]

$\bf \cfrac{18}{32}\implies \cfrac{2\cdot 3\cdot 3}{2\cdot 2\cdot 2\cdot 2\cdot 2}\implies \cfrac{3\cdot 3}{2\cdot 2\cdot 2\cdot 2}\implies \boxed{\cfrac{9}{16}}\qquad \checkmark \\\\[-0.35em] ~\dotfill\\\\ \cfrac{27}{48}\implies \cfrac{3\cdot 3\cdot 3}{2\cdot 2\cdot 2\cdot 2\cdot 3}\implies \cfrac{3\cdot 3}{2\cdot 2\cdot 2\cdot 2}\implies \boxed{\cfrac{9}{16}}\qquad \checkmark$

3 0

3 years ago

Find the value of this expression if x = 3<br> and y = -1.<br> xy²<br> 5

kap26 [50]

Answer:

<h2><em><u>0.6 = 6/10 = 3/5 is the answer.</u></em></h2>

Step-by-step explanation:

0.6 = 6/10 = 3/5

is the answer

This is because you have to substitute.

Given:

x = 3

y = -1

Unknown:

Final Answer

(x*y^2)/5

((3)(-1*-1)/5

= (3*1)/5

<h2><em><u>= 3/5</u></em></h2><h2><em><u>= 6/10</u></em></h2><h2><em><u>= 0.6</u></em></h2><h2><em><u></u></em></h2>

Hope this helped,

Kavitha

6 0

3 years ago

Encouraged by his investment, Joseph decides to open another CD account. This time, he deposits $1000 and plans to leave it unto

Lilit [14]

Answer:

1,113.87

Step-by-step explanation: Ps it was a joke man

5 0

3 years ago

Complete the table to describe the model. The model will have a verbal description, a division expression, and a fraction.

NARA [144]

Answer:

what I dont even know whT your talking about. could you please explain your question a bit more?? then maybe I could help you out

5 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