Help me plssssss!!!!!!!!!

Please help Anyone I'm seriously almost done

sp2606 [1]

Answer:

$\textsf{x = 45}$

Here, the given 2 angles are complementary angles. The sum is <u>90 degrees</u> as indicated by the right angle symbol.

$\textsf {Solving :}$

$\implies \textsf{x - 2 + 47 = 90}$

$\implies \textsf{x + 45 = 90}$

$\implies \textsf{x = 45}$

5 0

2 years ago

326482745445-453637+938464846-3845+784657

dmitriy555 [2]

The answer is going to be <span>327421537466</span>

4 0

4 years ago

Read 2 more answers

Help meeèeeeeeeeeeeee

mamaluj [8]

The answer is B your welcome

3 0

3 years ago

The triangle and rectangle

valentinak56 [21]

2x-4=3x-11 so add 4 to 11 and that’s -7 then subtract 2x from 3x so that’s 1x so 1x divides by -7 equals -7 so x=-7

7 0

3 years ago

According to a study conducted in one city, 39% of adults in the city have credit card debts of more than $2000. A simple random

Butoxors [25]

Answer:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean $\mu = 0.39$ and standard deviation $s = 0.0488$

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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 $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.39, n = 100$

Then

$s = \sqrt{\frac{0.39*0.61}{100}} = 0.0488$

By the Central Limit Theorem:

The sampling distribution of the sample proportion of adults who have credit card debts of more than $2000 is approximately normally distributed with mean $\mu = 0.39$ and standard deviation $s = 0.0488$

5 0

3 years ago