Area = (1/2) * 4 * 5 = 10 cm²
Answer:
288
Step-by-step explanation:
To solve, we need to find the area of the four triangle sides and add that to the square base's area.
Area of triangle sides: base * height * 0.5
Area of square base: base * height
Substitute the values
8 * 14 * 0.5 = 56
There are four of these sides, so the area of all the triangles is 224.
The square base is 8 * 8, or 64
Add these two values together.
64 + 224 = 288
Thus, the answer is 288
Answer: C
Step-by-step explanation:
To find the scale factor between circle P and circle Q, divide the diameter of circle Q by the diameter of circle P, 4.5 ft÷1.5 ft = 3. The scale factor between circle P and circle Q is 3.
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!