To write an improper fraction as a mixed number, divide the denominator into the numerator.
Image is provided.
Therefore, the improper fraction 11/9 can be written as the mixed number 1 and 2/9.
Answer:
20
Hope that helps!
Step-by-step explanation:
I think the answer is
A.) infinitely many
Answer:
<h3>The answer is 134.29 m</h3>
Step-by-step explanation:
First of all we need to convert 14.0 acres to m²
1.00 acre = 4046.86 m²
14.0 acres = 14 × 4046.86 = 56656.04 m²
Area of a circle = πr²
where
r is the radius
To find the radius substitute the value for the area into the above formula and solve for the radius
That's

Divide both sides by π
We have

Find the square root of both sides

r = 134.29139
r = 134.29 m to 2 decimal places
Hope this helps you
Answer:
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given that the mean of the Population = 95
Given that the standard deviation of the Population = 5
Let 'X' be the random variable in a normal distribution
Let X⁻ = 96.3
Given that the size 'n' = 84 monitors
<u><em>Step(ii):-</em></u>
<u><em>The Empirical rule</em></u>


Z = 2.383
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = P(Z≥2.383)
= 1- P( Z<2.383)
= 1-( 0.5 -+A(2.38))
= 0.5 - A(2.38)
= 0.5 -0.4913
= 0.0087
<u><em>Final answer:-</em></u>
The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors
P(X⁻ ≥ 96.3) = 0.0087