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3 years ago

Solve for t. 4=12gt^2

Mathematics

1 answer:

ivann1987 [24]3 years ago

6 0

Answer:

$t = \sqrt{ \frac{8}{g} }$

Step-by-step explanation:

We are solving for t.

$4 = \frac{1}{2} g {t}^{2}$

Multiply both sides by 2.

$8 = gt {}^{2}$

Divide both sides by

$\frac{8}{g} = {t}^{2}$

Take square root of both sides.

$t = \sqrt{ \frac{8}{g} }$

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4 years ago

Read 2 more answers

A box contains four red balls and eight black balls. Two balls are randomly chosen from the box, and are not

castortr0y [4]

P(B) = 8/12

P(R | B) = 4/11

P(B ∩ R) = 8/33

The probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

<em><u>Solution:</u></em>

<em><u>The probability is given as:</u></em>

$Probability = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}$

Given that,

A box contains four red balls and eight black balls

Red = 4

Black = 8

Total number of possible outcomes = 12

Let event B be choosing a black ball first and event R be choosing a red ball second.

$P(B) = \frac{8}{12}$

$P(B n R) = P(B) \times P(R)\\\\P(B n R) = \frac{8}{12} \times \frac{4}{11}\\\\P(B n R) = \frac{8}{33}$

<h3><u>Find </u><u> P(R | B)</u></h3><h3> $P(R | B) = \frac{P(R n B)}{P(B)}\\\\P(R | B) = \frac{\frac{8}{33}}{\frac{8}{12}}\\\\P(R | B) = \frac{8}{33} \times \frac{12}{8}\\\\P(R | B) = \frac{4}{11}$ </h3>

<em><u>The probability that the first ball chosen is black and the second ball chosen is red is about percent</u></em>

$\frac{8}{33} \times 100 = 0.24 \times 100 = 24 \%$

Thus the probability that the first ball chosen is black and the second ball chosen is red is about 24% percent

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4 years ago