3.75/ (1/16)=60 sixteenths =>
It’s the difference between the y values divided by two and subtracted from the bigger one. And the same thing for x values. U are kinda finding the second one
Answer is 279
Step-by-step explanation:
Lets do the following:
How many 1/9 are there in 1? Easy! 9. We can corroborate this by summing 1/9 nine times of multiplying 1/9 nine times. In both cases the result is 9.
Then, how many 1 are there in 31? Easier!! 31. Is we sum the number 1 thirty one times, the result is 31.
Now let join both results:
If each 1 has nine 1/9, and there are thirty one 1s in 31, then in 31 we have 31*9=279 1/9s. This is:
31*1 = 31 * 9 (1/9) = 279 * (1/9)
<h3>
Answer: Choice A) 0.20</h3>
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Explanation:
Let's say there are 1000 students. The students must take math, science or they can take both simultaneously.
- 65% of them are in math. So there are 0.65*1000 = 650 math students.
- 43% are in science, leading to 0.43*1000 = 430 science students.
- 13% are in both so we have 0.13*1000 = 130 students who are in both.
Now onto the sentence that says "Suppose a high school student who is enrolled in a math class is selected at random"
This means we only focus on the 650 math students and ignore the 1000-650 = 350 students who aren't in math.
Of those 650 math students, 130 are also in science (since 130 are in both classes).
The probability we're after is therefore 130/650 = 0.20
