Consider this expression Vx^4

How do u do 7.86 divided by 0.2

Kryger [21]

Answer:

39.3

Step-by-step explanation:

Did you need step by step or no?

4 0

4 years ago

The proportion of high school seniors who are married is 0.02. Suppose we take a random sample of 300 high school seniors; a.) F

cricket20 [7]

Answer:

a) Mean 6, standard deviation 2.42

b) 10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c) 14.85% probability that we find less than 4 of the seniors are married.

d) 99.77% probability that we find at least 1 of the seniors are married

Step-by-step explanation:

For each high school senior, there are only two possible outcomes. Either they are married, or they are not. The probability of a high school senior being married is independent from other high school seniors. 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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

In this problem, we have that:

$n = 300, p = 0.02$

a.) Find the mean and standard deviation of the sample count X who are married.

Mean

$E(X) = np = 300*0.02 = 6$

Standard deviation

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{300*0.02*0.98} = 2.42$

b.) What is the probability that, in our sample of 300, we find that 8 of the seniors are married?

This is P(X = 8).

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

$P(X = 8) = C_{300,8}.(0.02)^{8}.(0.98)^{292} = 0.1040$

10.40% probability that, in our sample of 300, we find that 8 of the seniors are married.

c.) What is the probability that we find less than 4 of the seniors are married?

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

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

$P(X = 0) = C_{300,0}.(0.02)^{0}.(0.98)^{300} = 0.0023$

$P(X = 1) = C_{300,1}.(0.02)^{1}.(0.98)^{299} = 0.0143$

$P(X = 2) = C_{300,2}.(0.02)^{2}.(0.98)^{298} = 0.0436$

$P(X = 3) = C_{300,3}.(0.02)^{3}.(0.98)^{297} = 0.0883$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0023 + 0.0143 + 0.0436 + 0.0883 = 0.1485$

14.85% probability that we find less than 4 of the seniors are married.

d.) What is the probability that we find at least 1 of the seniors are married?

Either no seniors are married, or at least 1 one is. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

From c), we have that $P(X = 0) = 0.0023$ . So

$0.0023 + P(X \geq 1) = 1$

$P(X \geq 1) = 0.9977$

99.77% probability that we find at least 1 of the seniors are married

3 0

3 years ago

Find the discount on a note for $23,000 made by Jacques on February 10 to be repaid on August 10 of the same year with a discoun

Olenka [21]

Total Value =$23,000

Time = 10 february to 10 august = 6 months

And

Rate of discount = 9% =0.09

Now we have to find the discount on the note

So we multiply the amount , time and rate

but time should be in years

So 6 months = 6/12 years = 0.5 years

So

Discount = 23,000 * 0.5 * 0.09 = 1035

Hence total Discount = $1035

3 0

4 years ago

Read 2 more answers

4m^2 + 4m + 1 = 0 quadratic formula

Alex73 [517]

X= -0.5. Get Desmos, it's really helpful for this sort of thing.

4 0

3 years ago

Point P was rotated about the Irgun (0,0) by -15

mart [117]

Answer:

D

Step-by-step explanation:

The point you're looking for is the same distance from the origin, but 15° clockwise around a circle centered at the origin. Points A and D are the only ones with the same radius. Point D is the only one that is 15° clockwise from P.

_____

The approximate coordinates of the rotated point are (-4.3120, 3.2259).

7 0

3 years ago

Consider this expression Vx^4 - y^2 help