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3 years ago

If a basketball was dropped 32 feet from a second floor of a home, how far would it rebound if it bounciness was

Mathematics

2 answers:

liubo4ka [24]3 years ago

7 0

23.68 percent I hope this help you!!!

valentina_108 [34]3 years ago

4 0

Answer:

23.68

Step-by-step explanation:

74% = 74/100

$\frac{74}{100}$ · $\frac{32}{1}$ = 23.68

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Convert 4x=1+2y from standard form to slope intercept form

gregori [183]

Answer:

3x=2y

Step-by-step explanation:

subtract on from each side 4x-1=1-1+2y=3x+2y

7 0

3 years ago

Please help 20 points and brainliest

Maksim231197 [3]

Answer: the last second one: 3/4 0.8 94%

Step-by-step explanation:

because 3/4 is smaller than one so it has to go first.

hope it helps :) pls give brainliest

7 0

2 years ago

Read 2 more answers

Factorize 9/16 x2 + xy + 4/9 y2

miskamm [114]

$\frac{9}{16} x^2 + xy + \frac{4}{9} y^2 = (\frac{3}{4} x)^{2} + xy + (\frac{2}{3}y)^{2} = (\frac{3}{4} x)^{2} + 2* \frac{3}{4} x * \frac{2}{3}y + (\frac{2}{3}y)^{2}$ $= ( \frac{3}{4}x + \frac{2}{3} y)^{2}$

7 0

2 years ago

Grumpy Corp drug tests all of the recent college graduates it hires each year. The drug test currently used correctly determines

Vedmedyk [2.9K]

Answer:

Hence, the probability of a Negative test is $P(T^C)=0.5904$ .

Step-by-step explanation:

(a)

Let $D=$ Drug user and $T=$ Test positive

Probability of drug users $P(D)=36\%=0.36$

$P(\frac{T}{D})=96\%=0.96$

$P(\frac{T^c}{D^c})=90\%=0.90$

(b)

Find the Probability of a Negative test,

$P(T^C)=P(\frac{T^C}{D^C})\cdot P(D^C)+P(\frac{T^C}{D})\cdot P(D)$

$\Rightarrow P(T^C)=0.9\cdot(1-0.36)+(1-0.96)\cdot 0.36$

$\Rightarrow P(T^C)=0.9\cdot 0.64+0.04\cdot 0.36$

$\Rightarrow P(T^C)=0.576+0.0144$

$\Rightarrow P(T^C)=0.5904$

4 0

2 years ago

Motorola used the normal distribution to determine the probability of defects and the number

vovangra [49]

Answer:

a.P<x<9.85 orx>10.15)=0.3174, Total defects=317.4

b.p=0.0026,total defects=2.6

c.Less of the items produced will be classified as defects.

Step-by-step explanation:

a.The standard score,z, is esentially x reduced by process mean then divided my process standard deviation.

$\mu=10,\sigma=0.15\\Therefore:-\\z=\frac{x-\mu}{\sigma}=\frac{9.85-10}{0.15}\approx-1.0\\z=\frac{x-\mu}{\sigma}=\frac{10.15-10}{0.15}\approx+1.0\\P(x10.15)=P(Z+1.0)\\=2P(Z$