Plese hlp brianly poepel.

What is the approximate lateral area of the cylinder? Use 3.14 for π and round to the nearest whole number.

Marysya12 [62]

108 cm is the correct number for the answer and the correct number is 339

3 0

3 years ago

Read 2 more answers

What is the value of the following:A)3/8*0.125+2.5

professor190 [17]

Answe

la respuesta es 3

Step-by-step explanation:

el 3/8 equivale a 0.375 sumando las 2 cantidades anteriores suman a 3

3 0

3 years ago

Gannon had to cut a

Colt1911 [192]

He would be able to cut 4 pieces.

7 0

3 years ago

If n ||m, find the value of x and the value of y.

babunello [35]

Answer: Option C

$y=5$ , $x=10$

Step-by-step explanation:

If the lines m and n are parallel then the large triangle and the small triangle are similar triangles, because they have equal angles.

By definition for similar triangles it is satisfied that if they have sides of length a, b, c and a ', b' and c 'then:

$\frac{a}{a'}=\frac{b}{b'} =\frac{c}{c'}$

In this case

$a = 12\\a' = 4\\c = 15\\c' = y$

So:

$\frac{12}{4}=\frac{15}{y}$

$3=\frac{15}{y}$

$y=\frac{15}{3}$

$y=5$

Therefore

if $x+y=15$ and $y=5$ then $x =10$

Finally

$y=5$ , $x=10$

5 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

4 years ago