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Elza [17]
2 years ago
13

Suppose the manager of a gas station monitors how many bags of ice he sells daily along with recording the highest temperature e

ach day during the summer. The data are plotted with temperature, in degrees Fahrenheit (°F), as the explanatory variable and the number of ice bags sold that day as the response variable. The least squares regression (LSR) line for the data is Yˆ = −114.05+2.17X. On one of the observed days, the temperature was 82 °F and 66 bags of ice were sold.
1. Determine the number of bags of ice predicted to be sold by the LSR line, Yˆ, when the temperature is 82 °F.
2. Compute the residual at this temperature.
Mathematics
1 answer:
Maslowich2 years ago
4 0

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

Equation of regression line :

Yˆ = −114.05+2.17X

X = Temperature in degrees Fahrenheit (°F)

Y = Number of bags of ice sold

On one of the observed days, the temperature was 82 °F and 66 bags of ice were sold.

X = 82°F ; Y = 66 bags of ice sold

1. Determine the number of bags of ice predicted to be sold by the LSR line, Yˆ, when the temperature is 82 °F.

X = 82°F

Yˆ = −114.05+2.17(82)

Y = - 114.05 + 177.94

Y = 63.89

Y = 64 bags

2. Compute the residual at this temperature.

Residual = Actual value - predicted value

Residual = 66 - 64 = 2 bags of ice

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Step-by-step explanation:

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s2008m [1.1K]

Answer:

C. -5 power of 5

Step-by-step explanation:

-5 power of 7 / -5 power of 2

Soo, you subtract the exponent 7 by 2 and you are left with 5.

You are thus left with -5 to the power of 5.

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Sidana [21]
AE = AC = 4

m<CAB = 60 (equilateral triangle)
m<CAE = 90 (square)

m<BAE = 150 (= 60 + 90)

Triangle BAE is isosceles since AB = AE;
therefore, m<AEB = m<ABE.

m<AEB + m<ABE + m<BAE = 180

m<AEB + m< ABE + 150 = 180

m<AEB + m<AEB = 30

m<AEB = 15

In triangle ABE, we know AE = AB = 4;
we also know m<BAE = 150, and m<AEB = 15.

We can use the law of sines to find BE.

BE/(sin 150) = 4/(sin 15)

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3 0
3 years ago
g In a one-tail hypothesis test where you reject H0 only in the lower tail, what is the p-value if Z STAT
zepelin [54]

Answer:

The P-value is 0.0166.

Step-by-step explanation:

<u>The complete question is:</u> In a​ one-tail hypothesis test where you reject H0 only in the lower ​tail, what is the p-value  if ZSTAT = -2.13.

We are given that the z-statistics value is -2.13 and we have to find the p-value.

Now, the p-value of the test statistics is given by the following condition;

                    P-value = P(Z < -2.13) = 1 - P(Z \leq 2.13)

                                  = 1 - 0.9834 = <u>0.0166</u>

Assuming that the level of significance is 0.10 or 10%.

The decision rule for rejecting the null hypothesis based on p-value is given by;

  • If the P-value of the test statistics is less than the level of significance, then we have sufficient evidence to reject the null hypothesis.
  • If the P-value of the test statistics is more than the level of significance, then we have insufficient evidence to reject the null hypothesis.

Here, the P-value is more than the level of significance as 0.0166 > 0.10, so we have insufficient evidence to reject the null hypothesis, so we fail to reject the null hypothesis.

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2 years ago
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