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3 years ago

Answer this due soon

Mathematics

2 answers:

likoan [24]3 years ago

5 0

Answer:

The base angles theorem converse states if two angles in a triangle are congruent, then the sides opposite those angles are also congruent. The Isosceles Triangle Theorem states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.

Step-by-step explanation:

IRINA_888 [86]3 years ago

5 0

I can’t see the picture 1245

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Use the table. What is the total amount o

irina1246 [14]

Answer:

1.89

Step-by-step explanation:

0.09+1.5+0.3

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3 years ago

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Complete the equation of the line through (6,−6) and (8,8)

vlabodo [156]

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3 years ago

Find the solution set of 5x<36 when x ∈ n

Margaret [11]

Answer:

5x<36

5x/5 < 36/5

x < 7.2

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3 years ago

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7nadin3 [17]

Answer: 54 liked best

Step-by-step explanation: None

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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

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

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3 years ago