Does anyone know this

PLEASE HELP , TRYING TO BE ON HONOR ROLL !

borishaifa [10]

Option A

The solution is $(0, \frac{-1}{4})$

Solution:

Given system of equations are:

3x + 6y = 1 ------ eqn 1

x - 4y = 1 ------ eqn 2

We have to find solution to system of equations

We can use substitution method

From eqn 2,

x = 1 + 4y -------- eqn 3

Substitute eqn 3 in eqn 1

3(1 + 4y) + 6y = 1

3 + 12y + 6y = 1

18y = 1 - 3

18y = -2

Divide both sides by 18

$y = \frac{-2}{18}\\\\y = \frac{-1}{4}$

Substitute the above value of y in eqn 3

$x = 1 + 4 \times \frac{-1}{4}\\\\x = 1-1\\\\x = 0$

Thus solution is $(0, \frac{-1}{4})$

8 0

3 years ago

At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine

Tresset [83]

Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $s = \frac{\sigma}{\sqrt{n}}$ .

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