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3 years ago

The table represents some points on the graph of a linear function.

Mathematics

1 answer:

Anika [276]3 years ago

3 0

Answer:

I'm sorry did you ever get the answer to this question

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Algebra 2 Please Help

hammer [34]

No sé lo que quieres decir

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3 years ago

What is the solution to this equation?

Bingel [31]

X=25 jjjjjjjjjjjjjjjjjjjjjjjjjjjj

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4 years ago

Read 2 more answers

Y-1/4=-3/8 solve for y

Leona [35]

Answer:

y=-1/8

Step-by-step explanation:

Add 1/4 to -3/8

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3 years ago

find the greatest number that exactly divide 16 and 24. ind the ornatost number that diido 20 and 20 min

laila [671]

Step-by-step explanation:

The greatest common factor (GCF) is the largest common factor that the numbers share. Here, 8 is the largest common factor of 16 and 24. So the GCF of 16 and 24 is 8.

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2 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