Jaimie is practicing for the long jump. Each week she makes 10 jumps and after each jump she records the distance, in meters, she reaches with each jump. Her results for this week are represented in the table below. Jump 1 2 3 4 5 6 7 8 9 10 Distance (m) 5.1 5.3 5.4 5.2 5.5 5.4 5.2 5.3 5.5 3.5 Jaimie then reviews her results and determines the mean and standard deviation so that she can compare her results from this week to her results from the previous week. Jaimie felt the 10th jump was not reflective of her ability because she stumbled before making the jump and decided to remove the 10th jump from her data set. Determine whether each statement describes the change that will occur to the mean and standard deviation from this week's results when Jaimie removes the 10th jump from her data set. Select Yes or No for each statement. Removing the 10th jump from her data set will cause her average jump distance to increase. Removing the 10th jump from her data set will cause her average jump distance to decrease.
The standard deviation will increase because the jump distances will be more spread out.
The standard deviation will decrease because the jump distances will be more clustered together.

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation:

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.

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step-by-step explanation:

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)