Find the measure of the missing angles.<br> 106e<br> d<br> 54

The organizers of a fair projected a 25 percent increase in attendance this year over that of last year, but attendance this yea

Gnom [1K]

Answer: 64 %

Step-by-step explanation:

Assume attendance last year to be equal to "x"

- An increase in percentage sees us add the percentage amount to 100.

- A decrease in percentage sees us subtract the percentage amount from 100.

Projected attendance = ((100+25)/100) * x = 1.25x

Actual attendance = ((100-20)/100) * x = 0.80x

Hence ratio of a actual to projected attendance = 0.80x/1.25x = 0.64

Convert into percentage = 0.64 * 100 = 64 %

5 0

3 years ago

PLZ HELP:

UkoKoshka [18]

1. In each graph, the two lines are parallel to each other.

2. Each set of equations have the same slope.

3. Equations that have the same slope as each other will always be parallel to each other.

8 0

3 years ago

A charity receives 2025 contributions. Contributions are assumed to be mutually independent and identically distributed with mea

uysha [10]

Answer:

The 90th percentile for the distribution of the total contributions is $6,342,525.

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