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3 years ago

Please help me solve this

Mathematics

1 answer:

Maurinko [17]3 years ago

6 0

Since AB and QR are equal set them equal and plug in your answer for x for AB

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A shop advertised a 22% off sale. If a coat originally sold for $255, find the decrease in price and sale price.

Umnica [9.8K]

It would be: 255 - (255 * 22%)
= 255 - (255 * 0.22)
= 255 - 56.1
= 198.9

In short, Decrease in Price is $56.1 & Sale Price is $198.9

Hope this helps!

8 0

3 years ago

I need help I need help I need help

vekshin1

Answer:

Since angle d, a, and e are supplementary, d+a+e =180 degrees.

Given that line LK and BC are parallel

By the alternate interior angle theorem, e=c and d=b.

Substitute:

d+a+e=b+a+c=180

Since b+a+c is the sum of the angles of triangle ABC, the sum of the interior angle of triangle ABC (which could be any triangle of any size and shape) is 180 degrees.

5 0

3 years ago

Please HELP mee !!! Chance flips a coin and rolls a 1-6 number cube. What is the probability that he rolls a number less then 4

nalin [4]

B. 20, hope this helps :)

8 0

4 years ago

If a = 1/2, what is the value of a²?

Ipatiy [6.2K]

Answer:

1/4

Step-by-step explanation:

a^2 = 1/2^2

1/2^2= 1/4

5 0

3 years ago

Read 2 more answers

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume

kobusy [5.1K]

Answer:

The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018

Step-by-step explanation:

The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula

$W = \frac{X - \mu}{\sigma} = \frac{X - 182}{6.5465}$

In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.

The cummulative distribution function of W will be denoted by $\phi$ . The values of $\phi$ can be found in the attached file.

$P(X > 201.0476) = P(\frac{X-182}{6.5465} > \frac{201.0476 - 182}{6.5465}) = P(W > 2.91) = 1-\phi(2.91) = \\1-0.9982 = 0.0018$

Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.

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6 0

3 years ago

Please help me solve this​

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