Mathematics, 07.03.2021 14:00 taylortayshaun7

66% of U. S. adults have very little confidence in newspapers. You randomly select 10 U. S. adults. Find the probability that the number of U. S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.

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a) 0.1434 = 14.34% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.7731 = 77.31% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.022 = 2.2% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. Adults are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

66% of U.S. adults have very little confidence in newspapers.

This means that p = 0.66

You randomly select 10 U.S. adults.

This means that n = 10

(a) exactly five

This is P(X = 5). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.66)^{5}.(0.34)^{5} = 0.1434$

0.1434 = 14.34% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.66)^{6}.(0.34)^{4} = 0.232$

$P(X = 7) = C_{10,7}.(0.66)^{7}.(0.34)^{3} = 0.2573$

$P(X = 8) = C_{10,8}.(0.66)^{8}.(0.34)^{2} = 0.1873$

$P(X = 9) = C_{10,9}.(0.66)^{9}.(0.34)^{1} = 0.0808$

$P(X = 10) = C_{10,10}.(0.66)^{10}.(0.34)^{0} = 0.0157$

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.232 + 0.2573 + 0.1873 + 0.0808 + 0.0157 = 0.7731$

0.7731 = 77.31% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

This is:

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.66)^{0}.(0.34)^{10} \approx 0$

$P(X = 1) = C_{10,1}.(0.66)^{1}.(0.34)^{9} = 0.0004$

$P(X = 2) = C_{10,2}.(0.66)^{2}.(0.34)^{8} = 0.0035$

$P(X = 3) = C_{10,3}.(0.66)^{3}.(0.34)^{7} = 0.0181$

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0 + 0.0004 + 0.0035 + 0.0181 = 0.022

0.022 = 2.2% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Answer from: Quest

its a straight line parallel to the time axis

Answer from: Quest

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584

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