A minor league plays 101 games,
They win 14 more than twice as many as they lost.
How many wins and loses?
Okay, simple.
First, write a simple addition equation. In this case wins = w and losses = l.
l + w = 101.
Now, we have to figure out a way to make one of the terms the same term as the other, in this case we can change the terms of w to l.
l = l
w = 2l +14 (14 more than twice the amount)
Okay. So plug in the new amount for w.
l + 2l + 14 = 101. Great! we now have a simple equation. lets solve.
put similar terms together.
3l + 14 = 101
3l = 87
l = 29
So, we have 29 loses, and
w = 2(29) + 14
72 wins!
To check,
72 + 29 = 101, correct!
Hope this helps!
Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
.
The probability of a month having a PCE between $575 and $790 is the <u>p-value of Z when X = 790 subtracted by the p-value of Z when X = 575</u>, hence:
X = 790:


Z = 1.8
Z = 1.8 has a p-value of 0.9641.
X = 575:


Z = -2.5
Z = -2.5 has a p-value of 0.0062.
0.9641 - 0.0062 = 0.9579.
0.9579 = 95.79% probability of a month having a PCE between $575 and $790.
More can be learned about the normal distribution at brainly.com/question/4079902
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10% of 30 is 3, which means it rained 3 days.
30-3 = 27 days NO rain
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