58​% 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. a) ​P(5)=
(b) ​P(x​≥6)=
​(c) ​P(x​<4)=

.216165788

.582225057

.417774943

Step-by-step explanation:

We need to use a binomial distribution here

A.

10C5*.58⁵*(1-.58)⁵= .216165788

B.

I honestly think the fastest way to solve this is adding the probabiblity of exactly 6,7,8,9,10

which means we write

10C6*.58⁶*(1-.58)⁴+10C7*.58⁷*(1-.58)³+10C8*.58⁸*(1-.58)²+10C9*.58⁹*(1-.58)+10C10*.58¹⁰= .582225057

C.

To solve this just take the compliment of answer B

1-.582225057= .417774943

here is the

mark brainliest pl

it will cost dollar $540

the living room is 3 which means if we divide 325/3 then we get $108

therefore if 1part of the living room cost $108 and 3 costed $325

then total 5part of the room will cost 5*108 which gives $540