Answer:
C y = 1/5x - 1
Step-by-step explanation:
Slope intercept form: y = mx + b where m = slope and b = y-intercept.
Substitute into the slope intercept form m= 1/5 and b = -1
y = 1/5x - 1
Answer: d
Step-by-step explanation:
Find the area of the rectangle.

Find the area of the base of the semicircle.

12 in would be the diameter, divide by 2 to get the radius.

Plug this into the formula.

Since we do not have a full circle but a semi one instead, we must divide our result by 2.

Add both areas.
180+56.52=236.52
To get to simplest form find the greatest common factor, or the gcf, of both numbers. To do that find all of the factors and find the one with the highest value both numbers share. So 30 would have factors of 1, 2, 3, 5, 6, 10, 15, and 30, and 42 would have the factors of 1, 2, 3, 6, 7, 14, 21, and 42. Both numbers share the factors of 1, 2, 3, and 6 so the gcf is 6. Now divide both numbers by six to get your answer.
Answer: 5/7
Answer:
Step-by-step explanation:
1. There are only 4 numbers including 5 that follow the "five or more" requirment, and the probability of spinning it once is 4/8, or 1/2. (The total sections is 8) Then we multiply 1/2 and 1/2 together to get the "two times in a row" requirement done. (1/2)*(1/2)= 1/4 is the probability.
2. There are two values on the spinner that are a multiple of 3, 3 itself and 6. Again, the total amount of numbers/sections is 8, so the probability of spinning a multiple of three is 2/8 or 1/4. The probability of spinning an odd number is 4/8 or 1/2. (1/2)*(1/4)=1/8 is the probability.
3. The probability of spinning one odd number is 1/2, and so we multiply 1/2 by itself four times. (1/2)*(1/2)*(1/2)*(1/2)=1/16 is the probability.
4. There are 6 numbers greater than two on the number wheel not including two itself. So the probability of that is 6/8, or 3/4. Then we multiply 3/4 by itself 3 times as it asks. (3/4)*(3/4)*(3/4)*(3/4)=81/256 is the probability.
Note that I am not really sure about the answer myself, but I hope that this can help in some way. Good luck! :)