18-3=15
Originally pay 25 = 75
450-75=375
375 + 75 = 450
C) 18 people were originally going on the trip and 15 are going now.
Idk about D
In this problem, the average is already given. In order to solve for the missing length of jump rope we have the following solution:
Let x = be the total length
X = 4(112)
X = 448
Therefore,
34m + 1m + 212m = 247m
448m - 247m = 201 length of the fourth jump rope
0.75 is equal to 6\8
5.8 is > 4\5
a decimal greater than 5\8 is 1.2
0.2 > 1\4
0.7>0.6
a decimal smaller than 6\8 is 0.01
The one that is not equal to -7\5 is -11\5
The answer is D
To find this, multiply 9.2 by the decimal form of the percent (4.75). The answer is 43.7
The probability that it also rained that day is to be considered as the 0.30 and the same is to be considered.
<h3>
What is probability?</h3>
The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.
The probability that the temperature is lower than 80°F and it rained can be measured by determining the number at the intersection of a temperature that less than 80°F and rain.
So, This number is 0.30.
Hence, we can say that it was less than 80°F on a given day, the probability that it also rained that day is 0.30.
To learn more about the probability from the given link:
brainly.com/question/18638636
The above question is incomplete.
The conditional relative frequency table was generated using data that compared the outside temperature each day to whether it rained that day. A 4-column table with 3 rows titled weather. The first column has no label with entries 80 degrees F, less than 80 degrees F, total. The second column is labeled rain with entries 0.35, 0.3, nearly equal to 0.33. The third column is labeled no rain with entries 0.65, 0.7, nearly equal to 0.67. The fourth column is labeled total with entries 1.0, 1.0, 1.0. Given that it was less than 80 degrees F on a given day, what is the probability that it also rained that day?
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