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charle [14.2K]
3 years ago
11

I need answers for all! thank you will mark brainlist!! please help me i’m begging you <3

Mathematics
1 answer:
Darina [25.2K]3 years ago
6 0

Step-by-step explanation:

y = 2/5 x - 9/5

when x = -1,

the value of y = 2/5(-1) - 9/5

= -2/5 - 9/5 = -11/5

when x = 0

=> y = 2/5(0) -9/5

= 0-9/5 = -9/5

when x=1

=> y = 2/5(1) - 9/5

= 2/5 - 9/5 = 7/5

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Dual-energy X-ray absorptiometry (DXA) is a technique for measuring bone health. One of the most common measures is total body b
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Answer:

\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095

-The sample is too small to make judgments about skewness or symmetry.

H0:\mu_{1}=\mu_{2}

H1:\mu_{1} \neq \mu_{2}

t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013

p_v =2*P(t_{(14)}

So the p value is a very high value and using any significance level for example \alpha=0.05, 0,1,0.15 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

Step-by-step explanation:

First we need to find the difference defined as:

(Operator 1 minus Operator 2)

d1=1.326-1.323=0.003      d2=1.337-1.322=0.015

d3=1.079-1.073=0.006     d4=1.229-1.233=-0.004

d5=0.936-0.934=0.002   d6=1.009-1.019=-0.01

d7=1.179-1.184=-0.005      d8=1.289-1.304=-0.015

Now we can calculate the mean of differences given by:

\bar d=\frac{\sum_{i=1}^n d_i}{n}=-0.001

And for the sample deviation we can use the following formula:

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}=0.0095

Describe the distribution of these differences using words. (which one is correct)

We can plot the distribution of the differences with the folowing code in R

differences<-c(0.003,0.015,0.006,-0.004,0.002,-0.01,-0.005,-0.015)

hist(differences)

And we got the image attached. And we can see that the distribution is right skewed but we don't have anough info to provide a conclusion with just 8 differnences.

-The sample is too small to make judgments about skewness or symmetry.

Use a significance test to examine the null hypothesis that the two operators have the same mean. Give the test statistic. (Round your answer to three decimal places.)

\bar X_{1}=1.173 represent the mean for the operator 1

\bar X_{2}=1.174 represent the mean for the operator 2

s_{1}=0.1506 represent the sample standard deviation for the operator 1

s_{2}=0.1495 represent the sample standard deviation for the operator 2

n_{1}=8 sample size for the operator 1

n_{2}=8 sample size for the operator 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

H0:\mu_{1}=\mu_{2}

H1:\mu_{1} \neq \mu_{2}

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) like this:

t=\frac{1.173-1.174}{\sqrt{\frac{0.1506^2}{8}+\frac{0.1495^2}{8}}}=-0.013

Statistical decision

For this case we don't have a significance level provided \alpha, but we can calculate the p value for this test. The first step is calculate the degrees of freedom, on this case:

df=n_{1}+n_{2}-2=8+8-2=14

Since is a bilateral test the p value would be:

p_v =2*P(t_{(14)}

So the p value is a very high value and using any significance level for example \alpha=0.05, 0,1,0.15 always p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and a we don't have a significant difference between the two means.

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Kayla and latasha work for a department store the amount of sales they made over the last year was collected monthly in the data
tresset_1 [31]

Based on the information represented by the boxplot ;

  • Latasha's lowest sale amount = 50

  • Kayla's median is between 200 and 300

  • Latasha has a greater spread due to higher IQR value

1.) <em><u>The Lowest amount of sale made by Latasha in one month </u></em>

  • The minimum value is denoted by the starting position of the lower whisker on a boxplot.

  • Lowest amount of sale made by Latasha = 50

2.) <em><u>50</u></em><em><u>%</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>sales</u></em><em><u> </u></em><em><u>made</u></em><em><u> </u></em><em><u>by</u></em><em><u> </u></em><em><u>Kayla</u></em><em><u> </u></em><em><u>:</u></em>

  • 50% of sales made marks the median value in a boxplot, it is denoted by the vertical line in between the box.

  • 50% of sales made by Kayla is between 200 and 300

  • With median sale value being 250

3.) <em><u>Spread</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>middle</u></em><em><u> </u></em><em><u>50</u></em><em><u>%</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>sales</u></em><em><u> </u></em><em><u>:</u></em>

  • The measure of spread of the middle 50% of a distribution on a boxplot is the Interquartile range (IQR) of the distribution

  • IQR = Upper Quartile (Q3) - Lower quartile(Q1)

<u>For Latasha</u> :

  • Q3 = 450 (Endpoint of the box)
  • Q1 = 150 (starting point of the box)

  • IQR = 450 - 150 = 300

<u>For</u><u> </u><u>Kayla</u><u> </u><u>:</u><u> </u>

  • Q3 = 375 (Endpoint of the box)
  • Q1 = 100 (starting point of the box)
  • IQR = 375 - 100 = 275

  • Since, Latasha's IQR is greater than Kayla's, then Latasha has a greater mid 50% spread than Kayla.

Learn more :brainly.com/question/24582786

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