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2 years ago

Help meee please !!!

Mathematics

2 answers:

Molodets [167]2 years ago

7 0

Shh shah ahahaiaia suasion

maksim [4K]2 years ago

7 0

Set up two equalities with the given info using x and y as the number of questions in both

X + Y = 24
5x + 3y = 100

Now take the first equation and solve for one variable. We will start with X

X = 24 - Y

Now we have a value for x (24-y). Plug this value for x into the second equation

5(24-y) + 3y = 100

Now solve for y

120 - 5y + 3y = 100
120 -2y = 100
-2y = -20
Y= 10

Now we have the exact y value. Plug this value into the second equation to solve for x

5x + 3(10) = 100
5x + 30 = 100
5x = 70
X = 14

Now check your work by plugging both the value for x and y into the first equation

14 + 10 = 24
24 = 24

X = 14
Y = 10

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At a concert, 20% of the audience members were teenagers. if the number of teenagers at a concert was 360, what was the total nu

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3 years ago

The bill for a meal at a restaurant before tax is $84.00. the sales tax rate is 7% and you want to leave a 20% tip based on the

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2 years ago

Help plz ty gl please

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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

2 years ago