You are in charge of finding a venue for a concert. With the average square feet per

Mathematics

1 answer:

stiks02 [169]3 years ago

3 0

Answer:

184,499 will fit, Yes

Step-by-step explanation:

Proportions

We'll use proportions to answer the question and find the number of students fit into a building of 300 ft x 300 ft. The area of the building is

$300\ in\times 300\ in=90000\ in^2$

We are given a factor of 2.05 students per square feet, thus the number of students that fit into the building is

$2.05\cdot 90000\ in^2=184500$

Since 184500 is greater than 142551 people required, they easily fit into that space.

The closest answer is

184,499 will fit, Yes

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Assume that adults have IQ scores that are normally distributed with a mean of mu equals 105 and a standard deviation sigma equa

Sindrei [870]

Answer:

0.7698

Step-by-step explanation:

If you call your random variable $X$ , then what you are looking for is

$P(87 \leq X \leq 123)$

because you want the probability of $X$ being between 87 and 123.

We need a table with of the normal distribution. But we can only find the table with $\mu = 0$ and $\sigma = 1$ . Because of that, first we need to normalize our random variable:

$Z = \frac{X - \mu}{\sigma} = \frac{X - 105}{15}$

(you can always normalize your variable following the same formula!)

now we can do something similar to our limits, to get a better expression:

$\frac{87 - 105}{15} = \frac{-18}{15} = -1.2$

$\frac{123 - 105}{15} = \frac{18}{15} = 1.2$

And we transform our problem to a simpler one:

$P(87 \leq X \leq 123) = P(-1.2 \leq Z \leq 1.2) = P(Z \leq 1.2) - P(Z \leq -1.2)$

(see Figure 1)

From our table we can see that $P(Z \leq 1.2) = 0.8849$ (this is represented in figure 2).

Remember that the whole area below the curve is exactly 1. So we can conclude that $P(Z \geq 1.2) = 0.1151$ (because 0.8849 + 0.1151 = 1). We also know the normal distribution is symmetric, then