Answer:
150.72 cm^2
Step-by-step explanation:
The blue area has an angle of 120 degrees, and 120/360 = 1/3, which shows that the blue area is 1/3 of the circle. In that case, the blue area is just 144pi/3. That simplifies to 48 pi.
48*3.14 = 150.72
Answer:
(2,2)
Step-by-step explanation:
<span>5(3x^3)^3-2(3x^3)
=</span><span>5(3^3 x^9) - 6x^3
= 135x^9 - 6x^3
= 3x^3 (45x^6 - 2)</span>
Answer:
They can invite 136 while staying in budget
Step-by-step explanation:
Firstly, you know they will have to pay an $80 cleaning fee, therefore, you can subtract 80 from 5,000:
5,000-80 = 4,920
After doing this, you can take 4,920 and divide it by 36 (the price per person).
4,920/36 = 136.67
Because you can not have .67 of a person, your closest number you can use is 136.
leaving them with $24
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01

X[bar] ± 
174.5 ± 
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!