How do I solve 321-207=342

gretchen and Ezia received equal scores on a test made up of multiple choice questions and an essay. Gretchen got 18 multiple ch

timurjin [86]

The multiply choice were worth 4 points.
Gretchen= 18x+19
Ezia= 15x+31

So basically you can enter any number. At first I picked 3 and then I did the math and I got 73 points for Gretchen and 76 points for Ezia.

Then I pick four and then I got 91 points for them both. This is how I got the answer:

18x+19=15x+31
18(4)+19=15(4)+31
72+19=60+31
91=91

5 0

3 years ago

Read 2 more answers

We have n = 100 many random variables Xi ’s, where the Xi ’s are independent and identically distributed Bernoulli random variab

777dan777 [17]

Answer:

(a) The distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b) The sampling distribution of the sample mean will be approximately normal.

(c) The value of $P(\bar X>0.50)$ is 0.50.

Step-by-step explanation:

It is provided that random variables $X_{i}$ are independent and identically distributed Bernoulli random variables with p = 0.50.

The random sample selected is of size, n = 100.

(a)

Theorem:

Let $X_{1},\ X_{2},\ X_{3},...\ X_{n}$ be independent Bernoulli random variables, each with parameter p, then the sum of of thee random variables, $X=X_{1}+X_{2}+X_{3}...+X_{n}$ is a Binomial random variable with parameter n and p.

Thus, the distribution of $X=\sum\limits^{n}_{i=1}{X_{i}}$ is a Binomial distribution.

(b)

According to the Central Limit Theorem if we have an unknown population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

The sample size is large, i.e. n = 100 > 30.

So, the sampling distribution of the sample mean will be approximately normal.

The mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu=p=0.50$