Interpret the slope in terms of the situation. Does it make sense?Interpret the y-intercept in terms of the situation. Does it make sense?

Determine the equation of the line of best fit. Write your answer in slope-intercept form. Any non-integers in this problem should be entered as decimal numbers. Round to three decimal places when necessary.

Answers

Answer 1

Answer:

a) Yes the slope makes sense, as the players foot length gets larger, their overall height also increases signifying that there is a correlation between foot size and height.

b) The y-intercept occurs when the line "intercepts" the y-axis. It also makes sense because a player that has 25 cm in distance from heel to toe may very likely be 1.55 meters in height. This is probable.

c) The equation of any line is:

y=mx+b (m= slope, b= y-intercept)

The slope of this line can be found by locating two points where the line intersects. I will use points (25.5,1.6) and (26.25,1.7).

The formula to find the slope is (y(2)-y(1)) / (x(2)-x(1)) = m

1.7 - 1.6 / 26.25 - 25.5 = .1/.75 = .13333

This line crosses the y-intercept at about 1.55m

The equation of this line is:

y = .1333 x + 1.55

Answer 2

Answer:

Answer:

y = 0.133x - 1.791

Step-by-step explanation:

The points (27, 1.8) and (27.75, 1.9) lie on the line of best fit. Use these points to find the slope:

m = y2 - y1 1.9 - 1.8 0.1

------------- = ----------------- = -------- = 0.133

x2 - x1 27.75 - 27 0.75

Plug the value of the slope into the equation of the line y = mx +b to get .

y = 0.133x + b

Substitute the point (27, 1.8) into the equation and solve for b:

1.8 = 0.133(27) + b

b = -1.791

Plug the value of b into the equation of the line of best fit to get:

y = 0.133x - 1.791

--------------------------------------------------------------------------------------------------------------------------

This answer came straight from the sample answer on Edmentum.com.

Related Questions

Right triangle ABC is shown. Which of these is equal to cos(A)?A) cos(B) B) cos(C) C) sin(B) D) sin(C)
Find the average value of f(x) = 7x^2(x^3+1)6 over the interval [0, 2].
Find the slope of the line that asses through the pair of oints (17,11) and (21,19)
Please Help Me For the figure below m<ABC = 70°. Which theorem or postulate can be used to determine m<DBC? What is m<DBC? (pic of the drawing below)
Find the volume of a cylinder, giventhe radius as 7cm and height 10cm.

Gregg has 12 cards. Half are black, and half are red. He picks two cards out of the deck. What is the probability that both cards are red

Answers

The solution is, the probability that both cards are red is 5/22.

What is probability?

Probability can be defined as the ratio of the number of favorableoutcomes to the total number of outcomes of an event.

here, we have,

given that,

Gregg has 12 cards.

Half are black, and half are red.

i.e. red = 6 & black = 6.

He picks two cards out of the deck.

now, we have,

Probability of a red card out of 12 cards = 6/12 = 1/2

Now, probability of a red card of 11 cards = 5/11

so, we get,

Combined probability = 1/2 * 5/11 = 5/22

Hence, The solution is, the probability that both cards are red is 5/22.

To learn more on probability click:

brainly.com/question/11234923

#SPJ3

Probability of a red card out of 12 cards = 6/12 = 1/2
Now, probability of a red card of 11 cards = 5/11

Combined probability = 1/2 * 5/11 = 5/22

In short, Your Answer would be 5/22

Hope this helps!

What is decimal representation of 27/100

Answers

0.27 is the decimal representation of 27/100.

.27 or divide 27/100 and you get the same answer

A certain test preparation course is designed to help students improve their scores on the MCAT exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 7 students' scores on the exam after completing the course: 37,12,12,17,13,32,23 Using these data, construct a 80% confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 1 of 4 : Calculate the sample mean for the given sample data. Round your answer to one decimal place.

Answers

Answer:

The 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

Step-by-step explanation:

The net change in 7 students' scores on the exam after completing the course are:

S = {37 ,12 ,12 ,17 ,13 ,32 ,23}

Compute the sample mean and sample standard deviation as follows:

$\bar x=(1)/(n)\sum x=(1)/(7)* 146=20.857\n\ns=\sqrt{(1)/(n-1)\sum (x-\bar x)^(2)}}=\sqrt{(1)/(7)* 622.8571}=10.189$

As the population standard deviation is not known, a t-interval will be formed.

Compute the critical value of t for 80% confidence interval and 6 degrees of freedom as follows:

$t_(\alpha/2, (n-1))=t_(0.20/2, (7-1))=t_(0.10,6)=1.415$

*Use a t-table.

Compute the 80% confidence interval for the average net change in a student's score after completing the course as follows:

$CI=\bar x\pm t_(\alpha/2, (n-1))*(s)/(√(n))$

$=20.857\pm 1.415*(10.189)/(√(7))\n\n =20.857\pm 5.4493\n\n=(15.4077, 26.3063)\n\n\approx (15.4,26.3)$

Thus, the 80% confidence interval for the average net change in a student's score after completing the course is (15.4, 26.3).

Can someone please help me –4x ≤ 36

Answers

Answer:

x > -9

Step-by-step explanation:

–4x ≤ 36

Divide by -4, remembering to flip the inequality

-4x/-4 > 36/-4

x > -9

The weight, in pounds, of a certain type of adult squirrel is normally distributed with a mean of 4.1 pounds and a standard deviation of 0.5 pounds. What percentage of squirrels have a weight between 3 and 5 pounds?

Answers

Answer:

95.02% of squirrels have a weight between 3 and 5 pounds.

Step-by-step explanation:

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.1 pounds

Standard Deviation, σ = 0.5 pounds

We are given that the distribution of weight is a bell shaped distribution that is a normal distribution.

Formula:

$z_(score) = \displaystyle(x-\mu)/(\sigma)$

P(weight between 3 and 5 pounds)

$P(3 \leq x \leq 5) = P(\displaystyle(3 - 4.1)/(0.5) \leq z \leq \displaystyle(5-4.1)/(0.5)) = P(-2.2 \leq z \leq 1.8)\n\n= P(z \leq 1.8) - P(z < -2.2)\n= 0.9641- 0.0139 = 0.9502= 95.02\%$

$P(3 \leq x \leq 5) = 95.02\%$

95.02% of squirrels have a weight between 3 and 5 pounds.

Final answer:

Using the concepts of normal distribution and Z-scores, it is found that approximately 95% of the squirrels have weights between 3 and 5 pounds.

Explanation:

The problem involves the use of the concept of normal distribution in statistics. In this case, we have a mean of 4.1 pounds and a standard deviation of 0.5 pounds. We have two weight points, namely 3 pounds and 5 pounds. To find out the percentage in this interval, we need to convert the weights into z-scores, which standardizes them. The formula for this is Z = (X - μ) / σ, where X is the weight point, μ is the mean, and σ is the standard deviation.

The Z-score for 3 pounds is Z1 = (3 - 4.1) / 0.5 = -2.2, and the Z-score for 5 pounds is Z2 = (5 - 4.1) / 0.5 = 1.8. Looking these Z-scores up in the standard normal table (or using a calculator with a normal distribution function), we get 0.9857 for Z1 and 0.9641 for Z2. The difference between these values gives the percentage of squirrels within the interval, which is 0.9641 - (1 - 0.9857) = 0.950 or 95%. Hence, approximately 95% of the squirrels have a weight between 3 and 5 pounds.