The answer would be t=11.6
Answer:
189.12 cm²
Step-by-step explanation:
If the radius of the stickers is 2 cm, then the diameter of each sticker is 4 cm.
1. Using this, divide each length of the sheet of paper by the diameter of 4 cm to see how many stickers can fit without overlapping.
24 ÷ 4 = 6 stickers
33 ÷ 4 = 8.25 = 8 stickers (only eight whole stickers will fit without overlap)
This means there can be a total of 48 stickers because 6*8 = 48.
2. Find the area of one sticker. The formula for the area of a circle if πr² (pi times radius squared) with 3.14 equal to π.
3.14 · 2² → 3.14 · 4 = 12.56 cm²
3. Multiply the area by the number of stickers that can fit on the sheet of paper (48 stickers).
12.56 · 48 = 602.88 cm²
4. Find the area of the sheet of paper. The formula for the area of a rectangle is length · width.
24 · 33 = 792 cm²
5. Subtract the area of the stickers from the area of the sheet of paper to solve for the remaining bare paper around the stickers.
792 - 602.88 = 189.12 cm²
Answer:
Concepts: Basic Linear Algebra
- X Cartons
- 1 Carton = 12 eggs
- Y Carton
- 1 Carton = 18 eggs
- 72 eggs total
- Hence 12X + 18Y = 72
- To meet your question guidelines:
- 12X+18Y-72=0
Answer:
Future value, A = $642
Step-by-step explanation:
Given the following data;
Principal = $500
Interest rate = 5% = 5/100 = 0.05
Time, t = 5 years
n = 365
To find the future value, we would use the compound interest formula;
Where;
A is the future value.
P is the principal or starting amount.
r is annual interest rate.
n is the number of times the interest is compounded in a year.
t is the number of years for the compound interest.
Substituting into the equation, we have;
Future value, A = $642
Answer:
a) Option C) The score was 2.49 standard deviations higher than the mean score in the class.
b) 2.3%
Step-by-step explanation:
a) We are given that the distribution of test grades is a bell shaped distribution that is a normal distribution.
Formula:


Option C) The score was 2.49 standard deviations higher than the mean score in the class.
b) The z-score for a particular score is -2.
We have to evaluate
P(z < -2)
Calculating the value from normal z-table.

Thus, 2.3% of of the class scored lower than my friend.