4) A right triangle has a hypotenuse length of 10 and one of its legs has a

Troyanec [42]

For x = -3, <span>(-0.7x) equals (-0.7[-3]), or 2.1.

8 8 8
Then f(-3) = ------------------- = ---------------- = ------------ = 0.31 (approx)
1 + 3*e^2.1 1 + 24.499 25.5</span>

5 0

3 years ago

What is the answer to negative 1/4-16 1/5 divided by 6???

Luba_88 [7]

For that last part you can leave your answer as a fraction of -59/20 or you can divide 29 by -59 and you’ll get -2.95.

4 0

3 years ago

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

Jayla recently accepted a new job earning $15.50 per hour. Last month, Jayla earned $2294 after four weeks of work. How many hou

V125BC [204]

Assuming the pay last month was from Jayla's new job, it represents pay for weekly hours of ...

$\dfrac{\$2294}{4\,week}\times\dfrac{1\,hours}{\$15.50}=37\,\dfrac{hours}{week}$

Jayla worked an average of 37 hours each week.

4 0

3 years ago

Explain how to use proportional reasoning to find 35% of 600

Sauron [17]

$\frac{35}{100} = \frac{x}{600}$

This is the equation we would use for this problem.

Now when we cross multiply, we get:
35(600)=100x

Simplify:
21000=100x

Divide each side by 100 to get x b itself:
210=x

So, as you can see, 210 is 35% of 600

8 0

3 years ago