-5х + Зу = -8<br> 3х - 7y = -3<br> Ordered pair

Victoria took a test and got 80% of the questions right.she answered 20 questions correctly how many questions were on the test?

lions [1.4K]

Answer:

d)25

Step-by-step explanation:

(80/100)*x=20

x=25

6 0

4 years ago

Each bag of skin contains 2.6 ft.³ of sand, how many bags of sand will need to be purchased to completely fill the sandbox?

steposvetlana [31]

Answer:skin?

Step-by-step explanation:

5 0

3 years ago

Find the values of x and y?<br> HELPPPPPPPP

ehidna [41]

x is 90 y is 43 im pretty sure

7 0

4 years ago

Suppose 46% of politicians are lawyers. If a random sample of size 662 is selected, what is the probability that the proportion

Svet_ta [14]

Answer:

0.9606 = 96.06% probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by less than 4%

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