What is the vertices

Jamal read a book with 128 pages then he read a book with 179 pages how pages did jamal read?

aliina [53]

128 pages+179 pages=307 pages

6 0

3 years ago

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What’s the difference in weight between the heaviest and lightest dog . Lassie 21 kg 249g Riley 23 kg 128g Fido 21,268g

Pie

Just Subtract Riley ( 23 kg 128g ) minus Lassie ( 21kg 249g ) and then you get your difference

4 0

3 years ago

50 POINTS GIVEAWAY!!!!!

Tema [17]

Answer:

1

Step-by-step explanation:

9 - 3/(1/3) + 1 = 1

4 0

3 years ago

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What is the solution to 3/4 (x+8) > 1/2 (2x+10)

densk [106]

Answer:

3/4x+2>x+5
3/4x-x>-8+5
-1/4x>-3
x>12

3 0

2 years ago

In the united states, voters who are neither democrat nor republican are called independent. it is believed that 12% of voters a

Nitella [24]

Answer:

a) 0.0245 = 2.45%

b) 0.9483 = 94.83%

c) 0.6931 = 69.31%

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are independent, or they are not. The probability of a person being independent is independent of any other person. So 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.

It is believed that 12% of voters are independent.

This means that $p = 0.12$

Survey of 29 people

This means that $n = 29$

a. what is the probability that none of the people are independent? probability =

This is P(X = 0). So

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

$P(X = 0) = C_{29,0}.(0.12)^{0}.(0.88)^{29} = 0.0245$

0.0245 = 2.45% probability that none of the people are independent

b. what is the probability that fewer than 7 are independent? probability =

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

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

$P(X = 0) = C_{29,0}.(0.12)^{0}.(0.88)^{29} = 0.0245$

$P(X = 1) = C_{29,1}.(0.12)^{1}.(0.88)^{28} = 0.0971$

$P(X = 2) = C_{29,2}.(0.12)^{2}.(0.88)^{27} = 0.1853$

$P(X = 3) = C_{29,3}.(0.12)^{3}.(0.88)^{26} = 0.2274$

$P(X = 4) = C_{29,4}.(0.12)^{4}.(0.88)^{25} = 0.2016$

$P(X = 5) = C_{29,5}.(0.12)^{5}.(0.88)^{24} = 0.1374$

$P(X = 6) = C_{29,6}.(0.12)^{6}.(0.88)^{23} = 0.0750$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0245 + 0.0971 + 0.1853 + 0.2274 + 0.2016 + 0.1374 + 0.0750 = 0.9483$