MATHEMATICS COLLEGE

Help please !!!! ASAP

Answers

Answer 1

Answer:

Answer: 1: cant figure out sorry :(

2: 2.5

3: 5

4: 20

5: 25

10: 50

Step-by-step explanation: What i did here was divided by 5 the numbers below 3 and multiply by 5 for the numbers above 3. Hope this helps :)

James has $9

Priya has $10.80

Siobhan has $9.18

***because 9*20% is 1.8 so add that to 9 to get 10.80 then times 10.80 by 15% to get 1.62 so you subtract 1.62 from 10.80 to get $9.18***

Solve for the area of ΔABC to the nearest whole number. A) 13 cm2 B) 26 cm2 C) 53 cm2 D) 106 cm2

Answers

Answer:

The answer is 53 cm2 or c.

To estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. If the population standard deviation is 8.2 years, computer the standard error of the mean. (Round to one decimal place) What is the probability that the sample mean age of the employees will be within 2 years of the population mean age

Answers

Answer:

The standard error of the mean is 1.3.

87.64% probability that the sample mean age of the employees will be within 2 years of the population mean age

Step-by-step explanation:

To solve this question, we have 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:

$Z = (X - \mu)/(\sigma)$

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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = (\sigma)/(√(n))$