I need help with this question!!!!!

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

4 0

The answer is D
y=1/4x-7

You might be interested in

Sixty percent of U.S. adults trust national newspapers to present the news fairly and accurately. You randomly select nine U.S.

Pavel [41]

Answer:

a) 0.251

b) 0.483

c) 0.0994

Step-by-step explanation:

We are given the following information:

We treat adult adults trusting national newspapers to present the news fairly and accurately as a success.

P(Adult trust) = 60% = 0.6

Then the number of adults follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 9

a) exactly five

$P(x = 15) = \binom{9}{5}(0.6)^{5}(1-0.6)^{9-5} = 0.2508 \approx 0.251$

b) at least six

We have to evaluate

$P(x \geq 6) = P(x = 6) + P(x = 7) + P(x=8) + P(x=9) \\= \binom{9}{6}(0.6)^6(1-0.6)^3 + \binom{9}{7}(0.6)^7(1-0.6)^2 + \binom{9}{8}(0.6)^8(1-0.6)^1 +\binom{9}{9}(0.6)^9(1-0.6)^0\\= 0.482609 \approx 0.483$

c) less than four

We have to evaluate

$P(x < 4) = P(x = 0) + P(x = 1) + P(x=2) + P(x=3) \\= \binom{9}{0}(0.6)^0(1-0.6)^9 + \binom{9}{1}(0.6)^1(1-0.6)^8 + \binom{9}{2}(0.6)^2(1-0.6)^7 +\binom{9}{3}(0.6)^3(1-0.6)^6\\= 0.099352 \approx 0.0994$