All categories

2 years ago

What is the area of the irregular figure below?

Mathematics

2 answers:

andrey2020 [161]2 years ago

8 0

Answer:

B, 48 inches

Step-by-step explanation:

First calculate the area of the parallelogram: 4 x 6

4 x 6 = 24

Then calculate the area of the triangle: 4 x 6

4 x 6 = 24

Then you add the two areas: (4 x 6) x 2 or 24 x 2

24 x 2 = 48

Therefore, your answer is B, 48 inches

damaskus [11]2 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

Which equation can be used to determine the value of x?

SSSSS [86.1K]

Answer:

choice 1) 2x + 40 = 180

Step-by-step explanation:

(x + 10) + (x + 30) = 180

2x+ 40 = 180

8 0

2 years ago

Read 2 more answers

Decrease 220kg by 12.5%.<br><br> NEED HELPPP!

Serga [27]

Hmm try 192.5 I think that’s right

6 0

3 years ago

Read 2 more answers

A cube has a volume of greater than 123 cm3. What are the possible lengths of the side of that cube?

horrorfan [7]

When we cube root 123, we get a decimal which rounds out to 4.97. So possible side lengths would be any number greater than that.<span />

4 0

3 years ago

A and b are odd numbers.

stepladder [879]

Answer:

2(3+7)=20 and 20 is a multiple of 4

Step-by-step explanation:

4 times 5 equals 20, which makes it a multiple of 4

3 and 7 are both odd, and added together get 10. Times that by 2 and you get 20

8 0

3 years ago

Read 2 more answers

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