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!
Answer:
Step-by-step explanation:
Concept 3: The dilation of a line segment is longer or shorter in the ratio given by the scale factor. If the scale factor is greater than 1 (k > 1), the image will be larger than the pre-image, making the segments (sides) of the image longer than the corresponding sides of its pre-image (an enlargement).
7 + 3.(2 - 3x) = 67
3 brackets are distributed
7+6-9x = 67
13-9x = 67
-9x = 67
x = 67/-9
2/7 as a fraction rounds to 0.3
0.3 * 3 = 0.9
3 x 2/7 is less than 3
You can't change more than 100 percent