Please answer : <br> (File below)

Find the perimeter area and area of the polygon.

Alika [10]

The area is 76 ft squared, and the perimeter is 43.1 ft. CORRECT ME IF IM WRONG BECAUSE I WAS SO BAD AT AREA/PERIMETER

4 0

3 years ago

At what corridnate point do y= 1/2 x + 4 and x + 4y = 4 intersect.

Viefleur [7K]

Answer:

(-4;2)

Step-by-step explanation:

1. if y=1/2 x+4, then 2y=x+8; it is possible to re-write as x-2y+8=0;

2. if x+4y=4, it is possible to re-write it as x+4y-4=0;

3. if x-2y+8=0 and x+4y-4=0, then x-2y+8=x+4y-4; ⇒ -2y+8=4y-4; ⇒ 6y=12; ⇔ y=2.

4. if y=2, then it is possible to substitute its value into any equation given in the condition:

x+4y=4 (substitution y=2); ⇒ x+8=4; ⇒ x= -4.

5. finally x= -4; y=2. It means (-4;2)

PS. the suggested solution is not the only way to resolve this task.

6 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

Safiya deposited money into a bank account that earned 6.5% simple interest each year. After 312 years, she had earned $36.40 in

kap26 [50]

The answer is 160.00

7 0

4 years ago

Read 2 more answers

Scott needs to buy a textbook for his college class. The book costs $221 new. He found a slightly used one online for $201. What

Artyom0805 [142]

Answer:

10 percent

Step-by-step explanation:

8 0

3 years ago

Please answer : (File below)