Mathematics, 17.07.2020 17:01 mimithurmond03

he data show the bug chirps per minute at different temperatures. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted temperature for a time when a bug is chirping at the rate of 3000 chirps per minute. Use a significance level of 0.05. What is wrong with this predicted value?

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Answer from: samone26

^Y= 26.72 + 0.0547Xi

^Y/ $_{X=3000}$ = 190.82ºF

B. It is unrealistically high.

Step-by-step explanation:

Hello!

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The data show the bug chirps per minute at different temperatures. Find the regression equation, letting the first variable be the independent (x) variable. Find the best predicted temperature for a time when a bug is chirping at the rate of 3000 chirps per minute. Use a significance level of .05. What is wrong with this predicted value?

Chirps in 1 min. 929 854 771 1004 1201 1027

Temperature (F) 81.3 77.3 64.8 80.3 92.2 80.9

What is the regression equation?

^y= _____ + _____

(Round the x-coefficient to four decimal places as needed. Round the constant to two decimals as needed)

What is the predicted value? ^y= _____ (Round to one decimal places as needed)

What is wrong with this predicted value?

A. The first variable should have been the dependent variable

B. It is unrealistically high.

C. It is only an approximation

D. Nothing is wrong with this value

To calculate the regression equation you have to estimate the slope and the y-intercept.

^Y= a + bX

Estimate of the slope:

$b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }$

n= 6

∑X= 5786 ∑X²= 5691944 $\frac{}{X}$ = 964.33

∑Y= 476.80 ∑Y²= 38277.76 $\frac{}{Y}$ = 79.47

∑XY= 465940.4

$b= \frac{465940.4-\frac{5786*476.80}{6} }{5691944-\frac{(5786)^2}{6} }= 0.0547$

Estimate of the Y-intercept:

$a= \frac{}{Y} -b*\frac{}{X}$

a= 79.47 -0.0547*964.33= 26.696= 26.72

The estimated regression equation is:

^Y= 26.72 + 0.0547Xi

^Y/ $_{X=3000}$ = 26.72 + 0.0547*3000= 190.82ºF

At the rate of 3000 chirps per minute it is expected a temperature of 190.82ºF

As you can see it is unrealistic to think that the chirping rate of bugs will have any effect over the temperature. For what is known about bugs, they tend to be more active to higher temperatures.

Considering the value obtained, as it is incredible high, if this regression was correct, every time the chirping rate of bugs increases, the ambient temperature would rise to levels incompatible with life.

I hope this helps!

Answer from: Quest

graph them on a graphing calculator, press 2nd trace. select intersect. hit enter until it says answer. ti 84 works best

step-by-step explanation:

Answer from: Quest

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + $\frac{15}{100}$ * x

4500-875 = $\frac{15}{100}$ * x

3625= $\frac{15}{100}$ * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x = $\frac{36250}{15}$

x= $2416.6

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