If PQ = 7 and PR = 32, find QR.* use diagram a.) 39 b.) 25 c.) 28 d.) 30

The time it takes to transmit a file always depends on the file size. Suppose you transmitted 30 files, with the average size of

Ostrovityanka [42]

Answer:

Y = 0.009042 + 0.0002457X

Y = 0.1073 seconds

Step-by-step explanation:

In the given problem we have two variables: the transmission time and average size of file.

Y = transmission time

X = average file size

The linear regression model is given by

Y = a + bX

The slope b is given by

b = correlation coefficient*(SDy/SDx)

Where SDy is the standard deviation of average transmittance time and SDx is the standard deviation of average file size.

b = 0.86(0.01/35)

b = 0.0002457

The y-intercept a is given by

a = y - bx

a = 0.04 - (0.0002457)126

a = 0.04 - 0.030958

a = 0.009042

Therefore, the linear regression model is

Y = 0.009042 + 0.0002457X

Predict the time it will take to transmit a 400 Kbyte file.

Substitute X = 400 in the regression model

Y = 0.009042 + 0.0002457(400)

Y = 0.009042 + 0.09828

Y = 0.1073 seconds

Therefore, the predicted time to transmit a 400 Kbyte file is 0.1073 seconds.

8 0

2 years ago

weekly wages at a certain factory are normally distributed with a mean of $400 and a standard deviation of $50. find the probabi

kenny6666 [7]

Answer:

0.34134

Step-by-step explanation:

In other to solve for this question, we would be using the z score formula

z = (x - μ) / σ

x = raw score

μ = mean

σ = Standard deviation

We are told in the question to find the probability that a worker selected at random makes between $350 and $400

let x1 = 350 and x2= 400 with the mean μ = 400 and standard deviation σ = $50.

z1 = (x1 - μ) / σ = (350-400) / 50 = -1

z2 = (x2 - μ) / σ = (400 - 400) / 50 = (0/50) = 0

From tables, P(z <= -1) = 0.15866

P(z <= 0) = 0.5

Then, the probability would give us, P(-1 ≤ z ≤ 0) =0.5 - 0.15866 =

0.34134

Hence, The probability that a worker selected at random makes between $350 and $400 = 0.34134

7 0

2 years ago

"A consultant compiled the following data set that shows" the number of visits made to the National Museum of American History f

aalyn [17]

Answer:

Step-by-step explanation:

The best option is for the consultant to remove these data points because they are outliers. Unusual data points which are located far from rest of the data points are known as outliers.

3 0

2 years ago

According to a report in USAToday, more and more parents are helping their young adult children get homes. Suppose eight persons

Inga [223]

Answer:

a) 95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.

( 0.0761 , 0.3239)

b) Margin of error = 0.1264.

Step-by-step explanation:

Explanation:-

Given '8' persons in a random sample of 40 young adults who recently purchased a home in Kentucky received help from their parents.

sample proportion of success $'p' = \frac{8}{40} = 0.2$

q = 1=p

q = 1-0.2 = 0.8

a)

95% confidence interval for the population proportion of all young adults in Kentucky who received help from their parents.

$(p-1.96\sqrt{\frac{pq}{n} } , p+1.96\sqrt{\frac{pq}{n} } )$

$(0.2-1.96\sqrt{\frac{0.2X0.8}{40} } , 0.2+1.96\sqrt{\frac{0.2X0.8}{40} } )$

(0.2 - 0.1239,0.2+0.1239)

( 0.0761 , 0.3239)

b) the margin of error for a 95% confidence interval for the population proportion.

For the 95% confidence interval ∝= 0.05 and zₐ = 1.96≅2.

$Margin of error = \frac{2\sqrt{pq} }{\sqrt{n} }$

$Margin of error = \frac{2\sqrt{0.2X0.8} }{\sqrt{40} }$

Margin of error for a 95% confidence interval for the population proportion.

Margin of error = 0.1264.

4 0

2 years ago

NEED HELP ON NUMBER 13 AND 14

Kazeer [188]

Answer:

Question 13: For age groups y=1 and y=1.3 response is 8 microseconds.

Question 14: The club was making a loss between 11.28 and 4.88 years.

Step-by-step explanation:

Question 13:

The age group y for which the response rate R is 8 microseconds is given by the solution of the equation

$8=y^4 +2y^3 - 4y^2 -5y +14.$

We graph this equation and find the solutions to be

$y=1;$ $y=1.302;$ $y=-2;$ $y=-2.302.$

Since only positive solutions for y are valid in the real world we take only those.

Thus only for age groups y=1 and y=1.3 the response is 8 microseconds.

Question 14:

The footbal club is making a loss when $p(t)$

Or

$t^3 -14t^2 +20t +120$

We graph this inequality and find the solutions to be

$t$ and $4.88$

Since in the real world only positive values for t are valid, we take the the second solution to be true.

Thus the club was making a loss in years $4.88$

5 0

2 years ago