Answer:
13/8 = 1 5/8
Step-by-step explanation:
first make 3 1/4 into an improper fraction
3 1/4 = (3*4+1)/4 = 13/4
13/4 /2
we know that dividing by a number is multiplying by its reciprocal, so
(13/4)/2 = 13/4 * (1/2)
multiply the numerators together, and multiply the denominators together
13/4 * (1/2) = (13*1)/(4*2) = 13/8 = 1 5/8
Answer:
it would be 36grms
Step-by-step explanation:
2 triangles makes a square
count how many sides in a cube
multi. how many triangles are in a side to how many sides
times how many triangles to how much grams each triangle weights
Answer:
270
Step-by-step explanation:
Because we do not know what the side lengths are, as long as they multiply to 30m^2, it's fine
For this question, let's just say the base is 5, and the height is 6. If we triple 5, we get 15, and if we triple 6, we get 18. 15*18=270
Now what if, the sides are not 5 and 6. Will the area still be the same? Let's find out.
10*3=30 so we can say for this answer, the base is 3, and the height is 10.
10 tripled is 30, and 3 tripled is 9. 30*9=270
So as we look at these 2 answers. we can conclude that the new area, no matter the side lengths, will be 270
Hope this helpes!
(1) -1.79, 2 1/4, 2.54 , -4
negative -> positive
reorder: -4. -1.79, 2 1/4 (or 2.25), 2.54
(2) -x
The larger x is the smaller the value
-$40.75<-$25.20
BUT the larger after the minus the more you owe so Renae owes more.
(3)
sorry idk
(4)
78F-x=70F
|x|=8*F
Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.
This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
1.9
Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.
My TI-83 calculator provided the following result:
normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)
