Find the area of the shaded region.

PLEASE HELP IS POSSIBLE :)

Arisa [49]

Answer: 9 inches

Step-by-step explanation:

3 0

3 years ago

Question 3 of 10 A Assume EFG = JKL. If mLJ = 90°, mL K = 26°, and m< L = 64°, what is the measure of < E?

kumpel [21]

The measure of angle E will be 90 degrees as well.

Assume EFG = JKL.

If mLJ = 90°, mL K = 26°, and m< L = 64°,

<h3>What is the congruent triangle?</h3>

Two triangles are said to be congruent if the length of the sides is equal, a measure of the angles are equal and they can be superimposed.

We have been given an image of two congruent triangles.

Since, EFG = JKL

therefore, the corresponding angles will be congruent.

m < J = 90° = m< E = 90

By congruence measure of angle, J will be equal to the measure of angle E.

Hence, The measure of angle E will be 90 degrees as well.

Learn more about congruent;

brainly.com/question/10586347

#SPJ1

4 0

2 years ago

308.3 divided by 15 but estimate the divisor and divided

aev [14]

Answer:

308.13/15=20.553 or 21

Step-by-step explanation:

hope this helps

plz mark brailiest

3 0

2 years ago

What is the best name to describe an object that has four right angles, the opposite sides are parallel, and it has four equal s

MAXImum [283]

Answer:

D-Square

Step-by-step explanation:

That describes a square, so the answer a square.

6 0

3 years ago

Read 2 more answers

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .

$39-55 \ \pm \ -2.015 \big{(}\sqrt{\frac{(8)^2}{13} +\frac{(2)^2}{18} } } \ \big{)}$

Simplify this expression.

$-16 \ \pm \ -2.015 (\sqrt{5.1453} \ )$
$-16 \ \pm \ 3.73139$

Adding and subtracting 3.73139 to and from -16 gives us a confidence interval of:

$(-20.5707,-11.4293)$

If we want to interpret the confidence interval of (-20.5707, -11.4293), we can say...

We are 90% confident that the interval from -20.5707 to -11.4293 holds the true mean of items sold at locations A and B.

5 0

1 year ago