What is the 5th term in the pattern 1,-4,9,-16?

Find the percent of the number.<br> 20% of 60 is

exis [7]

"of" means to multiply.
20% * 60
= 0.2 * 60 (since % means divide by 100)
= 12

3 0

3 years ago

Read 2 more answers

What is the radius of the circle given by the equation below?

Dmitry [639]

Answer:

C

Step-by-step explanation:

First, you gotta complete the square!

x^2+12x+36 = (x+6)^2

y^2-12y+36 = (y-6)^2

So, we have (x+12)^2 + (y-6)^2 = -56 + 36 + 36

The right side of the equation is the radius^2.

Simplify:

-56 + 36 + 36 = 16

The square root of 16 is 4, so the answer is C!

5 0

4 years ago

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

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