If you’re ADDING same bases, then you should also ADD the exponents. <br><br> True<br> False

Ashley needs 75 flower seeds for her garden. She wants to spend the least amount of money and have the fewest seeds left over. W

allochka39001 [22]

Not enough information to solve this problem

8 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

If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle,

liubo4ka [24]

Given:
Segment AC = 130 feet
Segment CD = 70 feet

I think that I'll be using the Pythagorean Theorem in finding the value of r. r will be the hypotenuse

Segment CE = (r - 70 feet)

r² = a² + b²
r² = 130² + (r-70)²
r² = 16,900 + (r-70)(r-70)
r² = 16,900 + r² - 70r - 70r + 4900
r² - r² + 140r = 16,900 + 4,900
140r = 21,800
r = 21,800/140
r = 155.71 feet

The radius of the circle is 155.71 feet.

8 0

3 years ago

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Did I put the decimal in the right place

il63 [147K]

Yes you did, you count up all the decimal spaces in all (which is 4 in this case) and put that amount of decimal spaces in the product.

5 0

3 years ago

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Jimmy has a cube labeled 1 through 6 A and a penny. How many different ways can the cube and penny land with one roll of the cub

PIT_PIT [208]

12. The penny can land two ways (heads or tails) and the cube can land six ways (1, 2, 3, 4, 5, or 6). Therefore, there are twelve possible results.