I need help ASAP! It's urgent.. PLISSSSS

A 45 degrees <br> B 90 degrees <br> C 135 degrees

drek231 [11]

Answer:

The measure of $\angle 3$ :

$\angle 3 = 45\textdegree$

Step-by-step explanation:

The measurement of $\angle 3$ is also $45\textdegree$ , because both $\angle 3$ and the angle with the measure of $45\textdegree$ are congruent.

4 0

3 years ago

The container that holds the water for the football team is 2/9 full. After pouring in 12 gallons of water, it is 2/3 full. How

Luden [163]

Answer:

27 gallons

Step-by-step explanation:

Our equation is: $\frac{2x}{9} +12 = \frac{2x}{3}$

What do we need to find: x

We need to first find x, so lets find x first.

$12 = \frac{2x}{3} - \frac{2x}{9}$

=> $12 = \frac{6x}{9} - \frac{2x}{9}$

=> $12 = \frac{4x}{9}$

=> x = 27

Lets go back to our equation.

$\frac{2(27)}{9} +12 = \frac{2(27)}{3}$

We are going to simplify this before solving this. We are going to get: $\frac{54}{9} +12 = \frac{54}{3}$

=> 18 = 18

Therefore, we found x and we checked that our x value is correct. Now, we need to find how many gallons can the container can hold.

$\frac{3x}{3} = a$ [a = gallons can the container hold]

=> x = a

Basically that means the container can hold 27 gallons of water.

Please give me brainliest if this is helpful ;D

4 0

2 years ago

Read 2 more answers

you start hiking at an elevation that is 80 meters below base camp. You increase your elevation by 42 meters. what is the new el

9966 [12]

<span>38 m below base camp.</span>

7 0

3 years ago

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where

z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704± $2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}}$ ≈ -0.171 ±0.237 that is (−0.408, 0.066)

7 0

3 years ago

Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

kondaur [170]

$f$ will be increasing on the intervals where $f'(x)>0$ and decreasing wherever $f'(x)$ . Local extrema occur when $f'(x)=0$ and the sign of $f'(x)$ changes to either side of that point.

$f'(x)$ is positive when $x$ is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that $f(x)$ is increasing on the intervals [-4, -2] and (2, 4].

$f'(x)$ is negative when $x$ is between some number between -2 and -1, up to some number less than 2. So $f(x)$ is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of $f'(x)$ changes for some $x$ between -2 and -1, and again to either side of $x=2$ .

4 0

3 years ago

I need help ASAP! It's urgent.. PLISSSSS ​

I need help ASAP! It's urgent.. PLISSSSS