Imaginary number of √-25 someone HELP

Mika and her two best friends are all dressing up as superheroes for Halloween this year. Their costumes are almost ready, but e

Sever21 [200]

Answer:

4yards

Step-by-step explanation:

4ft×3=12ft

1yrd=3ft

12ft÷3ft=4yards

6 0

3 years ago

5 is one-fourth of a number c

Gekata [30.6K]

Answer:

c=20

Step-by-step explanation:

5×4=20, so 5 is 1/4 of c. Hope that this helps you and have a great day :)

8 0

3 years ago

What is the value of x? 40 70 140 160 Answer: 70

Mnenie [13.5K]

Answer: 140
A triangle is 180 degrees in all

4 0

3 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 =

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

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

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

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

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

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

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

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

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

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