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kondor19780726 [428]
2 years ago
15

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:

36

Step-by-step explanation:

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Which equation can be used to determine the value of x?​
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choice 1) 2x + 40 = 180

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

2x+ 40 = 180

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Decrease 220kg by 12.5%.<br><br> NEED HELPPP!
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A cube has a volume of greater than 123 cm3. What are the possible lengths of the side of that cube?
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2 years ago
A and b are odd numbers.
stepladder [879]

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

Step-by-step explanation:

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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
5 0
2 years ago
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