Please help!!<br><br> Find the value of x and the length of each side

Solve for x: 2x - y = 3 x + y = 3?
? can someone explain this for me plz
8 Answers • Mathematics

Best Answer (Chosen by Voter)
Hi,

For this question, we are given the following system of equations:

2x - y = 3
x + y = 3

Let's make the first equation in the form of y = mx + b as shown below:

y = 2x - 3
x + y = 3

Now, substitute the first equation into the second to get:

x + 2x - 3 = 3

Combine similar terms to get:

3x = 6

x = 2

We now know that x = 2 and can plug this value into one of the given equations to get:

2 + y = 3

y = 1

FINAL ANSWER: x = 2 ; y = 1

I hope that helps you out!

7 0

3 years ago

Read 2 more answers

The mean amount purchased by a typical customer at Churchill's Grocery Store is $27.50 with a standard deviation of $7.00. Assum

Schach [20]

Answer:

a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.

b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00

c) 90% of sample means will occur between $26.1 and $28.9.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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 $s = \frac{\sigma}{\sqrt{n}}$ .