If parallel lines a and b are cut by transversal t, then which statement about angles 1 and 2 must be true?

Bassima evaluated the expression below.

amm1812

Answer:

The last one aka D is the answer on Edgenuity

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Researchers wanted to study physical activity in children, so they gave 74 randomly chosen children a "FitBit" to recorded how m

luda_lava [24]

Answer:

Step-by-step explanation:

The estimated regression equation to predict the daily steps taken is:

Daily step= 6350.02 - 217.90BMI + 2702.12Grade_middle + 601.44BMI * Grade_middle

where Grade_middle=1 if the student is in middle school, and Grade_middle=0 if the student is in high school

b) The slope of BMI is -217.90. The negative value of the slope indicates that BMI and Daily steps taken move in opposite direction.

We can say that for 1 unit increase in the BMI, the predicted value of Daily Steps taken by the student decreases by 217.90, while keeping other variables unchanged.

c) The regression equation for middle school students can be obtained by setting Grade_middle=1

Daily step= 6350.02 - 217.90BMI + 2702.12 * 1 + 601.44BMI * 1

= 6350.02 - 217.90BMI + 2702.12 + 601.44BMI

= 9052.14 + 383.54BMI

Therefore, the slope of BMI on steps taken for students in middle school is 383.54

The regression equation for high school students can be obtained by setting Grade_middle=0

Daily Steps = 6350.02 - 217.90BMI + 2702.12 x 0 +601.44BMI*0 = 6350.02 - 217.90BMI

Therefore, the slope of BMI on steps taken for students in high school is -217.90

The model says that variables, the daily steps taken and the BMI of a middle school student are positively correlated (move in the same direction) and the daily steps taken and the BMI are negatively correlated (move in opposite direction) for a high school student. The above slopes say that for 1 unit increase in the BMI for students in middle school increases the predicted daily steps taken by 383.54 and for 1 unit increase in the BMI for students in high school decreases the predicted daily steps taken by 217.90

d) The predicted value of daily steps taken for a BMI=20 and Grade_middle=1 is

Daily Steps = 6350.02 – 217.90 x 20 + 2702.12 x1 +601.44 x 20 x 1 = 16722.94

The actual value of the daily steps taken by a middle school student with a BMI of 20 is 7000

The residual is calculated as

$\text{residual}=\text{Actual}-\text{predicted}=7000-16722.94=-9722.94$

Therefore, this student’s residual is -9722.94

7 0

4 years ago

HELPP!!! 30 POINTS!!! WILL MARK BRAINLYIST!! Step 1: Written Response (30 points) Using complete sentences, compare the fat cont

yan [13]

Answer:

median a= 65 b= 75 c=68

This shows the median of each fat content of each restaurant showing that restaurant c is closer to that of 'restaurant a ' than a is to 'restaurant b' and near to 1/7 lower than that of restaurant b which shows they are all highly skewed with each other upon their graph.

The measures of spread start with 55 to 72 for 'restaurant a' where the upper and lower quartiles are 70-60 showing a distribution 10-17 into the whisker plot.

Compared to restaurant b which shows a measure of spread start with 65 to 90 for 'restaurant b' where the upper and lower quartiles are 85 - 68 into the whisker plot and show a spread of 3 in the left exterior and 5 spread in the right exterior to the upper quartile. So in comparison to restaurant a that restaurant b has a greater spread in the interquartile range = 15 where restaurant a =10.

For the outer quartile the comparison to restaurant a is twice as small meaning restaurant a is greater by over double and shown as 53 <a < 60 which when showing both outer quartiles we can use the amounts being 7 and 70<a<72 = 2

and show closed dots on equality line number line separately showing a<-7 and a>2 for each outer exterior quartile. For restaurant b this shows opposite similar equalities 65<b< 68 = b>-3 and 82<b<89 = b>7

So the differences are smaller for b for left side by 4 and larger for b by 5 up on the right side.

We compare both to restaurant c and find c = 60<c<61 = c>-1 and 70<c<71 = c>1 so the differences are that the outer quartile for c through the other quartiles = -2> c < -6 . This shows how smaller c is compared to a and b outer quartiles). We can also prove that while the outer quartiles are much more smaller for c than a and b we cna prove that c actually has an inner quartile more similar to c and closer distribution of b as the median is more closer for a and c where b has a greater output for median as restaurant b has the higher fat content and greater distribution within the inner quartiles over all.

Summary findings Restaurant c outer quartile is moderately skewed as they show -1 -0.5 on the left side and 1-0.5 on the right side. Restaurant a has a closer median inner quartile to c and closer distribution of inner distribution of c. It's output outer quartile distribiution is a distribution that is higher than b but a smaller inner quartile compared to b, when this happens then the distribution spread shows less range and fewer products to account for.

So i think restaurant b has the higher fat content to its menu.

Where restaurant a must be the healthiest as it holds the lowest range and larger gap is such range for healthier food.ie) when compared to the others.

Meanings of what we are asked.

In a box and whisker plot: the ends of the box are the upper and lower quartiles, so the box spans the interquartile range. the median is marked by a vertical line inside the box. the whiskers are the two lines outside the box that extend to the highest and lowest observations.

Skewness refers to distortion or asymmetry in a symmetrical bell curve, or normal distribution, in a set of data. If the curve is shifted to the left or to the right, it is said to be skewed. Skewness can be quantified as a representation of the extent to which a given distribution varies from a normal distribution.

As a general rule of thumb:

If skewness is less than -1 or greater than 1, the distribution is highly skewed.

If skewness is between -1 and -0.5 or between 0.5 and 1, the distribution is moderately skewed.

If skewness is between -0.5 and 0.5, the distribution is approximately symmetric.

4 0

4 years ago

HELP MEEEE I RLLY CANT KEEP DOING THIS

zloy xaker [14]

Answer:

y = -3x + 7

Step-by-step explanation:

7 0

3 years ago

PLEASE HELP ME IM SO DONE <br>Mean Value theorem to the f(x)=3e^(9x)-4x-9

MArishka [77]

((3e^9b-4b-9)-(3e^9a-4a-9))/(2)

7 0

3 years ago