x = perimeter
x<156
length = 66
so, in order to calculate perimeter you need to add two lengths and two widths
so
156 (perimeter) - 2 (66) = two widths
156 - 132 = 24 (remember this number is two widths added together)
so 24 twice the width SO 12 would be the number that the width can't be larger than
the width has to be less than 12
w < 12
Answer:
9
Step-by-step explanation:
4x9=36
5x9=45
6x9=54
7x9=63
(:
The solution to the equation r(1 - 2cosФ) = 1 is given as x² + y² - 4x√(x² + y²) + 4x² = 1
<h3>What is an
equation?</h3>
An equation is an expression that shows the relationship between two or more variables and numbers.
In polar form:
r = √(x² + y²) and cosФ = x / √(x² + y²)
Hence:
r(1 - 2cosФ) = 1
√(x² + y²) [1 - 2(x / √(x² + y²))] = 1
√(x² + y²) - 2x = 1
Take square of both sides:
x² + y² - 4x√(x² + y²) + 4x² = 1
The solution to the equation r(1 - 2cosФ) = 1 is given as x² + y² - 4x√(x² + y²) + 4x² = 1
Find out more on equation at: brainly.com/question/2972832
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The answer is A.
Add all the number values together and then divide it by the number of values you have: 43+32+65+88+50+72=350
350/6=58.33
Answer:
a) (1215, 1297)
b) (1174, 1338)
c) (1133, 1379)
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 1256
Standard Deviation, σ = 41
Empirical Rule:
- Also known as 68-95-99.7 rule.
- It states that almost all the data lies within three standard deviation for a normal data.
- About 68% of data lies within 1 standard deviation of mean.
- About 95% of data lies within two standard deviation of mean.
- About 99.7% of data lies within 3 standard deviation of mean.
a) range of years centered about the mean in which about 68% of the data lies

68% of data will be found between 1215 years and 1297 years.
b) range of years centered about the mean in which about 95% of the data lies

95% of data will be found between 1174 years and 1338 years.
c) range of years centered about the mean in which about all of the data lies

All of data will be found between 1133 years and 1379 years.