Answer:
Ethan can type 6 pages before meeting starts.
Step-by-step explanation:
Given that,
Ethan can type 2 pages in 1/8 hours
and his meeting is 3/4 hours late.
To find,
number of pages Ethan can type in 3/4 hours
1 hour = 60 minutes
1/8 hours = 1/8 * 60
= 7.5 minutes
3/4 hours = 3/4 * 60
= 45 minutes
7.5 minutes = 1 page typed
1 minute = 1/7.5 pages typed
45 minutes = (1/7.5) * 45
= 6 pages typed
1. 100 ABC, $16.25, $1,625, A. 97.68, B. $1,530.32
2. 100 DEF, $11.31, $1,131, A. $67.80, B. $1,062.20
3. 40 GHI, $9.15, $366, A. $21.96, B. $344.04
4. 100 JKL, $15.27, $1,527, A. $91.62, B. $1435.38
5. 100 MNO, $13.22, $1,322, A. $79.32, B. $1242.68
Hope This Helps!!!
a) The amount of concern is ...
(total interest)/(number of payments) = interest/payment
$14,644.95/120 ≈ $122.04 . . . . amount of interest per payment
__
b) The ratio of concern is ...
(total interest)/(total payments) × 100% = 14,644.95/39,644.95 × 100%
≈ 36.94% . . . . percent of total payments that is interest
Answer: 3 miles per hour
<u>Step-by-step explanation:</u>
Use the formula "distance (d) = rate (r) x time (t)" to create a system of equations.
Let "r" represent the rate they are rowing
Let "c" represent the current
time rate distance <u>EQUATION</u>
Downstream: 4 hours r + c 40 miles 4(r + c) = 40
Upstream: 10 hours r - c 40 miles 10(r - c) = 40
Distribute, then eliminate r to solve for c:
Down: 4r + 4c = 40 → 5(4r + 4c = 40) → 20r + 20c = 200
Up: 10r - 10c = 40 → -2(10r - 10c = 40) → <u>-20r + 20c</u> =<u> -80</u>
40c = 120
<u> ÷40 </u> <u>÷40 </u>
c = 3
We have strong evidence that on average, students study less than 150 minutes per night during the school week
Normal distribution:
mean μ₀ = 150
Sample:
Sample size n = 272
Sample mean x = 141
Sample standard deviation s = 66
The standard error of the sample mean SE = σ /√n
SE = 66/√272
SE = 66 / 16,49
SE = 4
Test Hypothesis:
Null hypothesis H₀ x = μ₀
Alternative hypothesis Hₐ x < μ₀
z(s) test statistics is:
z(s) = ( x - μ₀ ) / s/√n
z(s) = - 9 /4
z(s) = - 2,25
p-value for that z(s) p-value = 0,0122
Then for α = 0,05 p-value < 0,05
We are in the rejection region we need to reject H₀