What is the area of each piece of the composite figure?

Brilliant_brown [7]

Answer:

16.6 units

Step-by-step explanation:

Hi there!

We can use the Pythagorean theorem to help us solve this problem: $a^2+b^2=c^2$ where c is the longest side of a right triangle (the hypotenuse) and a and b are the other two sides

It's safe to assume that x is the longest side in this triangle, making it the c value. Plug 14 and 9 into the equation as a and b and solve for x:

$14^2+9^2=x^2\\196+81=x^2\\277=x^2\\x=16.6$

Therefore, the value of x is 16.6 units when rounded to the nearest tenth.

I hope this helps!

6 0

3 years ago

A tire manufacturer is considering a newly designed tread pattern for its all-weather tires. Tests have indicated that these tir

Nataly_w [17]

Answer:

With outlier (102)

$t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415$

$p_v =P(t_{9}>0.415)=0.344$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 125 feet at 5% of significance.

Without outlier (102)

$t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285$

$p_v =P(t_{8}>3.285)=0.0056$

And we conclude that we reject the null hypothesis since $p_v$ . So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier, removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Data: 129, 128, 130, 132, 135, 123, 102, 125, 128, 130

We can calculate the mean with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=126.2$ represent the sample mean

$s=9.138$ represent the sample standard deviation

n=10 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is significantly higher than 125, the system of hypothesis would be:

Null hypothesis: $\mu \leq 125$

Alternative hypothesis: $\mu > 125$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=10-1= 9$

Then since is a right tailed sided test the p value would be:

$p_v =P(t_{9}>0.415)=0.344$

Conclusion

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significantly higher than 125 feet at 5% of significance.

Using all the data we see that we don;t have enough info to conclude that the true mean is higher than 125, but if we see careful 9 of the 10 values are over the limit of 125 feet. And if we repeat the procedure with the outlier of 102. We got this:

$\bar X=128.89$ represent the sample mean

$s=3.551$ represent the sample standard deviation

$t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285$

First we need to calculate the degrees of freedom given by:

$df=n-1=9-1= 8$

Then since is a right tailed sided test the p value would be:

$p_v =P(t_{8}>3.285)=0.0056$

And we conclude that we reject the null hypothesis since $p_v$ . So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

8 0

3 years ago

Rosy has 16 pencils and 24 erasers. She is putting together packets that will havean equal number of pencils and erasers in each

N76 [4]

$16 = 2^4$

$24 = 2^3 \times 3$

$\text{HCF =}2^3 = 8$

Maximum number of packets = 8

16 ÷ 8 = 2
There are 2 pencils in each packet.

24÷8 = 3
There are 3 erasers in each bag.

Rosy can have a maximum of 8 packets. Each packet has 2 pencils and 3 erasers.

5 0

4 years ago

Read 2 more answers

You move right 3 units and left 4 units. you end at (-3,-1). where did you start?

Tema [17]

Answer:

(-2,-1) is where u start

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

For example( and this is straight from the paper), "15 people in 3 equal rows."

densk [106]

At a race, there are 3 rows with 15 people in each row. How many total people are there?

Numerical expression- 15 times 3

(Idk if this helped very much but it's worth a shot)

6 0

3 years ago