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Ann [662]
3 years ago
13

What error did she make? She simplified the denominator incorrectly. The denominator simplifies to –7. She labeled the points in

correctly. The point (–7, 4) should be (x1, y1). She used an incorrect formula. The formula should be the change in y-values with respect to the change in the x-values. She used an incorrect formula. The formula should be the sum of the x-values with respect to the sum of the y-values.

Mathematics
2 answers:
Svet_ta [14]3 years ago
6 0

Answer:

The formula should be the change in y-values with respect to the change in the x-values. She used an incorrect formula

Step-by-step explanation:

The question is incomplete. Here is the complete question.

Anya found the slope of the line that passes through the points (–7, 4) and (2, –3). Her work is shown below. Let (x2, y2) be (–7, 4) and (x1, y1) be (2, –3). m = = = The slope is . What error did she make? She simplified the denominator incorrectly. The denominator simplifies to –7. She labeled the points incorrectly. The point (–7, 4) should be (x1, y1). She used an incorrect formula. The formula should be the change in y-values with respect to the change in the x-values. She used an incorrect formula. The formula should be the sum of the x-values with respect to the sum of the y-values.

Slope of a line is expressed according to the formula;

Slope m = Δy/Δx

Slope = y2-y1/x2-x1

Given the coordinates (–7, 4) and (2, –3), from the coordinates;

x1 = -7, y1 = 4, x2 = 3, y2 = -3

Substitute into the formula;

Slope = -3-4/3-(-7)

Slope = -7/3+7

Slope = -7/10

Based on Anya workings, it can be concluded that Anya used an incorrect formula. The formula should be the change in y-values with respect to the change in the x-values

Mariulka [41]3 years ago
4 0

Answer:

Step-by-step explanation:

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The variance for the data is  17,507. 5.

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The weekly salaries of a sample of employees at the local bank are given in the table below.

Employee Weekly Salary Anja $245 Raz $300 Natalie $325 Mic $465 Paul $100.

<h3>Variance</h3>

Variance is the expected value of the squared variation of a random variable from its mean value, in probability and statistics.

The mean value of the salaries of employees is;

\rm Mean= \dfrac{245+300+325+465+100}{5}\\\\Mean=287

The variance is given by;

\rm  Variance=\sqrt{\dfrac{(245-287)^2 +(300-287)^2 +(325-287)^2 +(465-287)^2 +(100-287)^2}{5-2}} \\\\Variance = 132.316}\\\\On \ Squaring \ both \ the \ sides\\\\Variance^2=17,507. 5

Hence, the variance for the data is  17,507. 5.

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The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let µ denote the true average re
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Answer:

a) df=n-1=16-1=15

The statistic calculated is given by t=3.3  

Since is a one-side upper test the p value would be:      

p_v =P(t_{15}>3.3)=0.0024  

So since the p value is lower than the significance level pv we reject the null hypothesis.

b) df=n-1=8-1=7

The statistic calculated is given by t=1.8  

Since is a one-side upper test the p value would be:      

p_v =P(t_{7}>1.8)=0.057  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

c) df=n-1=26-1=25

The statistic calculated is given by t=-0.6  

Since is a one-side upper test the p value would be:      

p_v =P(t_{25}>-0.6)=0.723  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation      

\bar X represent the sample mean

s represent the standard deviation for the sample

n sample size      

\mu_o =20 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

t would represent the statistic (variable of interest)      

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.      

We need to conduct a hypothesis in order to determine if the true mean is higher than 20, the system of hypothesis would be:      

Null hypothesis:\mu \leq 20      

Alternative hypothesis:\mu > 20      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

(a) n = 16, t = 3.3, a = 0.05, P-value =

First we need to calculate the degrees of freedom given by:  

df=n-1=16-1=15

The statistic calculated is given by t=3.3  

Since is a one-side upper test the p value would be:      

p_v =P(t_{15}>3.3)=0.0024  

So since the p value is lower than the significance level pv we reject the null hypothesis.

(b) n = 8, t = 1.8, a = 0.01, P-value =

First we need to calculate the degrees of freedom given by:  

df=n-1=8-1=7

The statistic calculated is given by t=1.8  

Since is a one-side upper test the p value would be:      

p_v =P(t_{7}>1.8)=0.057  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

(c) n = 26,t = -0.6, P-value =

 First we need to calculate the degrees of freedom given by:  

df=n-1=26-1=25

The statistic calculated is given by t=-0.6  

Since is a one-side upper test the p value would be:      

p_v =P(t_{25}>-0.6)=0.723  

So since the p value is higher than the significance level pv>\alpha we FAIL to reject the null hypothesis.

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Answer:

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