Help 16-18 thank you

Mathematics

1 answer:

saw5 [17]3 years ago

5 0

Answer:

16. (80)(60) = 4800

17. (500)(600) = 300000

18. (700)(800) = 560000 a). high

Step-by-step explanation:

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(20+80 divided by 2x8) divided [(54 divided by 9+14)divided by 4]

lesya692 [45]

Answer:

Step-by-step explanation:

Let's write out the equation:

8(20+80/2)/[(54/9+14)/4]

For the first part we get 8(100/2)

Then 50x8

Then 400

For the second part we have (6+14)/4 as we divide 54 by 9. PEMDAS

Then 20/4

Then 5

So our final answer is 400/5 which is 80.

Hope this helps!

7 0

3 years ago

20. If a certain linear function f(x) is evaluated at 0, the answer is 4.

bija089 [108]

Answer:

y = (1/3)x + 4

Step-by-step explanation:

Two points on this line are (0, 4) and (3, 5).

As we move from the first point to the second, x increases by 3 and y increases by 1. Thus, the slope, m, of the line is m = rise / run = 1/3.

Use the slope-intercept equation: y = mx + b.

If we use the data from the point (0, 4), we get:

4 = (1/3)(0) + b, so that b = 4. The desired equation is y = (1/3)x + 4.

7 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

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