LOL
Answer:
D. The over 30s have a larger range and interquartile range than the under 30s
Step-by-step explanation:
In a data set, the range is the difference between the maximum and minimum. So, the range for under 30s is 20, while the range for over 30s is 24. Additionally, the interquartile range is the difference between Q3 and Q1. For a boxplot, Q3 is the line where the box ends and Q1 is the line where the box begins. Therefore, the IQR for the under 30s is 8, and the IQR for over 30s is 11. So, D must be correct.
The equation that represents that line is
y=2/3x+5.
First, you have to find the slope (the change of y over the change of x which here is 2/3. Then, you put one of the points in and solve. I used (3,7) When you input them into the equation, you get 7=2/3 (3)+_ Which gives you 7=2+_ .
5+2 is seven, so the blank is 5. So, when you do that, you get Y=2/3x+5.
If you need more of an explanation, comment. The answer is the second one
Hopes this helps :)
option C is the correct answer. sin β = 15 / 17
<h3>How to work it</h3>
<u>Given data</u>
Right angled triangle has sides
opposite = 15
adjacent = 8
Hypotenuse = 17
sin β = opposite / hypotenuse
substituting the values gives
sin β = 15 / 17
Read more on SOH CAH TOA here: brainly.com/question/20020671
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33 = p -6.71
To solve for P, add 6.71 to each side:
p = 33 + 6.71
p = 39.71
The second choice is correct.
Answer:
probability of a crash with at least one fatality if a driver drives while legally intoxicated (BAC greater than 0.09) = 0.001932
Step-by-step explanation:
P(BAC=0|Crash with fatality)=0.625
P(BAC is between .01 and .09|Crash with fatality)=0.302
P(BAC is greater than .09|Crash with fatality)=0.069
Let the event of BAC = 0 be X
Let the event of BAC between 0.01 and 0.09 be Y
Let the event of BAC greater than 0.09 be Z
Let the event of a crash with at least one fatality = C
P(X|C) = 0.625
P(Y|C) = 0.302
P(Z|C) = 0.069
P(C) = 0.028
probability of a crash with at least one fatality if a driver drives while legally intoxicated (BAC greater than 0.09) = P(C n Z)
But note that the conditional probability of probability that a driver is intoxicated (BAC greater than 0.09) given that there was a crash that involved at least a fatality is given by
P(Z|C) = P(Z n C)/P(C)
P(Z n C) = P(Z|C) × P(C) = 0.069 × 0.028 = 0.001932
Hope this Helps!!!
