Need help with this math

Mathematics

1 answer:

dybincka [34]4 years ago

7 0

It's negative four, you can see the line moves down by four. The y- intercept is 4 and the x- intercept is 1. Now to find the slope you use stardard form y = mx + b. once substituting the values in the equation you find out that it's -4.

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3 years ago

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>

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