Find the measure of the indicated angle.

List the 5 ways we can show that two triangles are congruent.

ser-zykov [4K]

Answer:

If two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent. If any two angles and the included side are the same in both triangles, then the triangles are congruent.

(side, side, side) we have two triangles with all three sides equal. ...

(side, angle, side) ...

(angle, side, angle) ...

(angle, angle, side) ...

(hypotenuse, leg)

Step-by-step explanation:

this is crazy lol

6 0

3 years ago

The cable company takes an SRS of 250 of the approximately 15,000 households who subscribe to them. They found that 75, percent

Damm [24]

Answer:

The mean of the sampling distribution of p is of 0.7 and the standard deviation 0.029

Step-by-step explanation:

We use the Central Limit Theorem to solve this question. So

Central Limit Theorem

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Percentage of 70%, sample of 250.

This means that $p = 0.7, n = 250$

Mean:

$\mu = p = 0.7$

Standard deviation:

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.7*0.3}{250}} = 0.0290$

The mean of the sampling distribution of p is of 0.7 and the standard deviation 0.029

5 0

3 years ago

Read 2 more answers

John is twice as old as Mary. The sum of their ages is 21. How old is Mary?

lesya692 [45]

J=john
m=mary
j is 2 times as old as mary
j=2 times m

sum is 21
m+j=21
find mary
subsitute

j=2m therefor subsitute 2m for j
m+2m=21
add like terms
3m=21
divide by 3
m=7

mary=7

*HOPE I HELPED!*

6 0

4 years ago

What is 3 1/2 x 2 1/3

Gemiola [76]

Answer:

8 1/6

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

If sin theta = 2/5 and theta is in quadrant I, determine the following.

anzhelika [568]

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{2}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a \qquad \begin{cases} c=\stackrel{hypotenuse}{5}\\ a=adjacent\\ b=\stackrel{opposite}{2}\\ \end{cases} \\\\\\ \pm\sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a\implies \stackrel{I~Quadrant}{+\sqrt{21}=a}$

recall that cosine is positive on the I Quadrant, so though we get a ± valid roots, only the positive one applies.

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{hypotenuse}{5}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}} \qquad \qquad cot(\theta)=\cfrac{\stackrel{adjacent}{\sqrt{21}}}{\stackrel{opposite}{2}} \\\\\\ csc(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{opposite}{2}} \qquad \qquad sec(\theta)=\cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}$

now, for tangent and secant, let's rationalize the denominator.

$\bf tan(\theta)\implies \cfrac{\stackrel{opposite}{2}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{21} \\\\\\ sec(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\stackrel{adjacent}{\sqrt{21}}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{5\sqrt{21}}{21}$

5 0

3 years ago