What is x divided by 6+x=

SOEMONE PLEASE TELL ME DID I CLICK THE RIGHT ONE PLEASE PLEASE HELP ME

Advocard [28]

Answer:

Yes.

Step-by-step explanation:

I had a problem just like this and you are correct.

8 0

2 years ago

n a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of inches and a stand

kotykmax [81]

Answer:

(a) The probability that a study participant has a height that is less than 67 inches is 0.4013.

(b) The probability that a study participant has a height that is between 67 and 71 inches is 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is 0.0401.

(d) The event in part (c) is an unusual event.

Step-by-step explanation:

The complete question is: In a survey of a group of men, the heights in the 20-29 age group were normally distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts (a) through (d) below. (a) Find the probability that a study participant has a height that is less than 67 inches. The probability that the study participant selected at random is less than inches tall is nothing. (Round to four decimal places as needed.) (b) Find the probability that a study participant has a height that is between 67 and 71 inches. The probability that the study participant selected at random is between and inches tall is nothing. (Round to four decimal places as needed.) (c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than inches tall is nothing. (Round to four decimal places as needed.) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

We are given that the heights in the 20-29 age group were normally distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches.

Let X = the heights of men in the 20-29 age group

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean height = 67.5 inches

$\sigma$ = standard deviation = 2 inches

So, X ~ Normal( $\mu=67.5, \sigma^{2}=2^{2}$ )

(a) The probability that a study participant has a height that is less than 67 inches is given by = P(X < 67 inches)

P(X < 67 inches) = P( $\frac{X-\mu}{\sigma}$ < $\frac{67-67.5}{2}$ ) = P(Z < -0.25) = 1 - P(Z $\leq$ 0.25)

= 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 0.25 in the z table which has an area of 0.5987.

(b) The probability that a study participant has a height that is between 67 and 71 inches is given by = P(67 inches < X < 71 inches)

P(67 inches < X < 71 inches) = P(X < 71 inches) - P(X $\leq$ 67 inches)

P(X < 71 inches) = P( $\frac{X-\mu}{\sigma}$ < $\frac{71-67.5}{2}$ ) = P(Z < 1.75) = 0.9599

P(X $\leq$ 67 inches) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{67-67.5}{2}$ ) = P(Z $\leq$ -0.25) = 1 - P(Z < 0.25)

= 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 1.75 and x = 0.25 in the z table which has an area of 0.9599 and 0.5987 respectively.

Therefore, P(67 inches < X < 71 inches) = 0.9599 - 0.4013 = 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is given by = P(X > 71 inches)

P(X > 71 inches) = P( $\frac{X-\mu}{\sigma}$ > $\frac{71-67.5}{2}$ ) = P(Z > 1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.9599 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.9599.

(d) The event in part (c) is an unusual event because the probability that a study participant has a height that is more than 71 inches is less than 0.05.

7 0

3 years ago

A Car can Travel 82 miles on 2 gallons of gasoline. How Far can it travel on 9 gallons?

Tom [10]

Answer:

369 miles

Step-by-step explanation:

82 miles =2 gallons

41 miles = 1 gallon

369 miles = 9 gallons

7 0

3 years ago