What is the product

In circle R, RD(line over it) is a radius and DF (line over it) is a tangent segment which statement must be true?

butalik [34]

We'll have an attached picture (Ignore the labeling). We can easily observe that DF (in the picture BC), cannot be equal to the radius. Then, we cannot also claim that RD=RF. And since we are talking about the tangent segment, we'll have a right triangle. The correct answer is C)

5 0

2 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

2 years ago

If V = 12R / (r + R) , then R =

irina1246 [14]

$V=\frac{12R}{r+R}\\\\V(r+R)=12R\\\\Vr+VR=12R\\\\VR-12R=-Vr\\\\(V-12)R=-Vr\ \ \ \ /:(V-12)\neq0\\\\R=\frac{-Vr}{V-12}\\\\R=\frac{Vr}{12-V}$