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3 years ago

9

Plz help PLZ IF CORRECT I’LL GIVE BRAINLIEST

1 answer:

vlada-n [284]3 years ago

8 0

Step-by-step explanation:

products of

$2 \frac{2}{3} \times 3\frac{3}{4} \\ = \frac{2 \times 3 + 2}{3} \times \frac{3 \times 4 + 3}{4} \\ = \frac{8}{3} \times \frac{15}{4} \\ = 10$

Hope it will help :)

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Engineers want to design seats in commercial aircraft so that they are wide enough to fit 95%of all males. (Accommodating 100%

Anarel [89]

Answer:

Upper P95 = 16.21in

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

$\mu = 14.4, \sigma = 1.1$

Upper P 95

This is the 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.

Then

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 14.4}{1.1}$

$X - 14.4 = 1.1*1.645$

$X = 16.21$

Upper P95 = 16.21in

3 0

3 years ago

What two numbers add to 92 and multiply to 160?

KatRina [158]

None But, 80 x 2 = 160 and 80 + 2 = 82, are you sure you mean 92?

3 0

3 years ago

How do I solve <img src="https://tex.z-dn.net/?f=sec%20-%202%5Cpi" id="TexFormula1" title="sec - 2\pi" alt="sec - 2\pi" align="a

olga2289 [7]

Answer:

Yes, buddy ut's 1 your correct x

6 0

3 years ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

3 years ago

HELP PLEASE !!! :)))))

zlopas [31]

Answer: the person has high blood sugar Levels

Step-by-step explanation:

8 0

3 years ago