Answer:
30,058 spectator

Explanation:
The total number of spectators is equal to the sum of West Stanford's and North Storm's supporters.
We are given that:
Total number of spectators = <span>71,167 spectator
North Storm spectators = </span><span>41,109 spectator
So, to get the number of West Stanford spectators, all we have to do is subtract North Storm spectators from the total spectators as follows:
West Stanford spectators = </span>71,167 - 41,109 = 30,058 spectator

Hope this helps :)

8 0

2 years ago

Evaluate. 58−(14)2=58-142= ________

nasty-shy [4]

For this case we have the following expression:

$58- (14) ^ 2$

The first step is to solve the quadratic term.

We have then:

$58- (14) ^ 2 = 58-196$

Then, the second step is to subtract both resulting numbers:

$58- (14) ^ 2 = -138$

We observe that the result obtained is a negative number.

Answer:

The result of the expression is given by:

$58- (14) ^ 2 = -138$

7 0

2 years ago

Vicki has 210 baseball cards and buys an additional 15 cards every week. Jeb has 275 baseball cards and buys an additional 10 ca

Hitman42 [59]

Answer:

11 weeks

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Management is considering adopting a bonus system to increase production. One suggestion is to pay a bonus on the highest 5 perc

Olegator [25]

Answer:

The bonus will be paid on at least 4099 units.

Step-by-step explanation:

Problems of normally distributed samples can be 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

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

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 percentile of this measure.

In this problem, we have that:

The highest 5 percent is the 95th percentile.

Past records indicate that, on the average, 4,000 units of a small assembly are produced during a week. The distribution of the weekly production is approximately normally distributed with a standard deviation of 60 units. This means that $\mu = 4000, \sigma = 60$ .