All categories

3 years ago

I need help with this question. It's about inequalities.

Mathematics

1 answer:

PtichkaEL [24]3 years ago

5 0

N is greater than it equal to 3

So you put a shaded in for a three and draw a line to the right

I think :)

You might be interested in

Solve 2 | 2x + 10|+ 4 > 10

Valentin [98]

Answer:

x>−72 or x<−132

Step-by-step explanation:

I believe that is the answer

8 0

3 years ago

Number one and number two

Amanda [17]

1. sum of exterior angles = 360

so one interior angles = 180 - (360/30) = 180 - 12 = 168 degrees
sum of all the interior angles = 168*30 = 5040
B

2. H

the sum of exterior angles of all polygons is 360 degrres

4 0

3 years ago

25.7 is what percent of 141?

Lady_Fox [76]

Answer:

about 18.2%

Step-by-step explanation:

All you have to do is divide the result of the percent, so 25.7, by the total, 141.

So 25.7÷141=0.18226950354

We round it to 0.182 and multiply it by 100 to know the percent which is 18.2

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

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

5 0

2 years ago

Jason eats 10 ounces of candy in 5 days. a. How many pounds does he eat per day?

Nadusha1986 [10]

A. He eats 1/8 a pound of candy each day.
B. It would take him a total of 8 days to eat a pound of candy.

A. Since he ate 10 ounces in 5 days, you divide 10 by 5 to find he ate 2 ounces per day. Since there are 16 ounces in a pound that would be your denominator (bottom of fraction) and 2 would be the numerator (top of your fraction). You will get 2/16 which if you simplify (divide by 2) you get 1/8.

B. Knowing that there are 16 ounces per pound and he ate 1/8 pound ( 2/16 pound or 2 ounces) a day, you can either look at the simplified denominator to get the answer, or, divide 16 by 2 to get your answer.

I hope this helps!

8 0

2 years ago

I need help with this question. It's about inequalities. ​

I need help with this question. It's about inequalities.