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3 years ago

Answer this fast please

Mathematics

1 answer:

Hatshy [7]3 years ago

7 0

Answer:

(-2,1)

Step-by-step explanation:

Rectangle has same lengths and same width so its (-2,1)

x= -2 y=1

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Mrs. Wright bought some notebook paper for her class. She decided to keep 3/4 of the paper for her own use. How much paper did s

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So the total amount of paper is 4/4 right?
And she kept 3/4.
So we have to minus 3/4 from 4/4.
Because it has the same denominator we need not change it to similar one.So 4-3 is one.
So answer is 1/4

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3 years ago

Need steps written and solved

BaLLatris [955]

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3 years ago

Should i do my teast for 12th grade calculus

Salsk061 [2.6K]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Is x=5 the solution to the following equation? Show your work with the substitution property

guapka [62]

Answer:

yes

Step-by-step explanation:

2x-4=6

Let x = 5

2*5 - 4 = 6

10 -4 =6

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3 years ago

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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

Answer this fast please​

Answer this fast please