Which statement is true about the purpose of trend lines in scatterplots?

How many quarters make a whole

Eduardwww [97]

Answer:

4

Step-by-step explanation:

4×(1/4) = 1

⁄(⁄ ⁄•⁄ω⁄•⁄ ⁄)⁄

3 0

3 years ago

Li’s recipe uses 3 2/3 cups of flour. He plans to make 4 batches. How much flour will li need. I need help on finding out what t

pishuonlain [190]

Answer:

14 and 2/3rd cups of flour for 4 batches

Step-by-step explanation:

3 0

3 years ago

Sandra writes (4x30) + (4x2) and says the quotient for 128 + 4 is 8. is she correct? explain.

Rashid [163]

No, because the equation shows her answer. 30 and 2 combined equals 32, which is the answer. 32x4=128 and 128/4=32

6 0

3 years ago

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One winter day, the outside temperature, y, was never below −1°. The inside of a car was always warmer than the temperature outs

mr Goodwill [35]

Answer:

It's C

Step-by-step explanation:

Your graph should be this one, with both lines solid because, the X line, or red, has to be at least -1 or higher, and the Y line, or the blue, has to be +4 or higher

4 0

3 years ago

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Quantitative noninvasive techniques are needed for routinely assessing symptoms of peripheral neuropathies, such as carpal tunne

Pavlova-9 [17]

Answer:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

$p_v =P(t_{(15)}>1.507)=0.076$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X_{CTS}=2.35$ represent the mean for the sample CTS

$\bar X_{N}=1.83$ represent the mean for the sample Normal

$s_{CTS}=0.88$ represent the sample standard deviation for the sample of CTS

$s_{N}=0.54$ represent the sample standard deviation for the sample of Normal

$n_{CTS}=10$ sample size selected for the CTS

$n_{N}=7$ sample size selected for the Normal

$\alpha=0.01$ 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 check if the mean for the group CTS is higher than the mean for the Normal, the system of hypothesis would be:

Null hypothesis: $\mu_{CTS} \leq \mu_{N}$

Alternative hypothesis: $\mu_{CTS} > \mu_{N}$

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:

$t=\frac{\bar X_{CTS}-\bar X_{N}}{\sqrt{\frac{s^2_{CTS}}{n_{CTS}}+\frac{s^2_{N}}{n_{N}}}}$ (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) the info given like this:

$t=\frac{2.35-1.83}{\sqrt{\frac{0.88^2}{10}+\frac{0.54^2}{7}}}}=1.507$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{CTS}+n_{N}-2=10+7-2=15$

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(15)}>1.507)=0.076$

We can use the following excel code to calculate the p value in Excel:"=1-T.DIST(1.507,15,TRUE)"

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the group CTS, NOT have a significant higher mean compared to the Normal group at 1% of significance.

6 0

3 years ago