PLEASEEEEE HELP ASAPPPPPP

On each of five slips of paper, you wrote the words eggs, waffles, pancakes, fruit, and muffin. Every day, you choose one slip f

vovangra [49]

1/5 is your answer. I hope I'm right
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4 0

3 years ago

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What is -6x-12+8y+16x simplified? and -4+5x-17-22x simplified?

tekilochka [14]

First bring like terms together before simplifying

6 0

3 years ago

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53% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

faltersainse [42]

Answer:

a

$P(X = 5) = 0.2423$

b

$P( X \ge 6 )= 0.4516$

c

$P( X < 4 )= 0.12694$

Step-by-step explanation:

From the question we are told that

The population proportion is p = 0.53

The sample size is n = 10

Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution

i.e

$X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial distribution is

$P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

Here C stands for combination hence we are going to be making use of the combination function in our calculators

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly five is mathematically represented as

$P(X = 5) = ^{10}C_5 * 0.53^5 * (1- 0.53)^{10-5}$

=> $P(X = 5) = 252 * 0.04182 * 0.023$

=> $P(X = 5) = 0.2423$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is at least six is mathematically represented as

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

=> $P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]$

=> $P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]$

=> $P( X \ge 6 )= 0.4516$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is less than four is mathematically represented as

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

=> $P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]$