just use what you know about this stuff
(a+36d)/(a+20d) = (a+55d)/(a+36d)
(a+36d)^2 = (a+55d)(a+20d)
a^2+72ad+1296d^2 = a^2+75ad+1100d^2
3ad = 196d^2
3a = 196d
That is, for any value of n,
a=196n
d=3n
So, there is no unique solution.
If n=1, then a=196 and d=3. The terms are
196+20*3 = 256
196+36*3 = 304
196+55*3 = 361
304/256 = 361/304
You can easily verify that it works for any value of n.
Answer:
59 to 66
Step-by-step explanation:
Mean test scores = u = 74.2
Standard Deviation =
= 9.6
According to the given data, following is the range of grades:
Grade A: 85% to 100%
Grade B: 55% to 85%
Grade C: 19% to 55%
Grade D: 6% to 19%
Grade F: 0% to 6%
So, the grade D will be given to the students from 6% to 19% scores. We can convert these percentages to numerical limits using the z scores. First we need to to identify the corresponding z scores of these limits.
6% to 19% in decimal form would be 0.06 to 0.19. Corresponding z score for 0.06 is -1.56 and that for 0.19 is -0.88 (From the z table)
The formula for z score is:

For z = -1.56, we get:

For z = -0.88, we get:

Therefore, a numerical limits for a D grade would be from 59 to 66 (rounded to nearest whole numbers)
Hello!
The formula for volume of a rectangular prism is:
V = lwh
Let's convert the dimensions to decimals, and then use the formula to solve.
8 1/2 = 8.5 as a decimal.
7 = 7 as a decimal.
2 1/2 = 2.5 as a decimal.
Multiply:
V = 8.5 × 7 × 2.5
V = 59.5 × 2.5
V = 148.75
148.75 = 148 3/4 as a fraction because if you convert 3/4 to a decimal by dividing the numerator by the denominator, it equals 0.75.
ANSWER:
The volume of the book is 148 3/4 inches cubed.
Answer:
Problem 1) frequency: 160 heartbeats per minute, period= 0.00625 minutes (or 0.375 seconds)
Problem 2) Runner B has the smallest period
Problem 3) The sound propagates faster via a solid than via air, then the sound of the train will arrive faster via the rails.
Step-by-step explanation:
The frequency of the football player is 160 heartbeats per minute.
The period is (using the equation you showed above):

second problem:
Runner A does 200 loops in 60 minutes so his frequency is:
loops per minute
then the period is: 0.3 minutes (does one loop in 0.3 minutes)
the other runner does 200 loops in 65 minutes, so his frequency is:
loops per minute
then the period is:

Therefore runner B has the smaller period
Answer:
See explanation.
Step-by-step explanation:

Common factor: y
