What is 110% expressed as a fraction? 55/5 110%/1 11/10

Freddy exercised 2.5 hours per 4 days. He burned 330 calories per hour. How many calories did Freddy burn by exercising?

Leto [7]

multiply

$330 \times 2 \\ 660 \times 4 \\ then \: the \: answer \: plus \\ \: 330 \div 2 \\ 165 \times 2$

that will be the answe

2964 calories

8 0

3 years ago

Read 2 more answers

If 15 is 25% of a value, what is that value? A. 60 B. 65 C. 40 D. 75

Bess [88]

It’s 60..................

3 0

3 years ago

What is the domain of the function y=2 x-5?<br> O x>-5<br> O x>2<br> O x>5<br> O x>10

antiseptic1488 [7]

That’s a good one Because what is the domain of the function Y equals 2X

3 0

4 years ago

Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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 normally distributed samples are solved using 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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$