The residents of a small town and the surrounding area are divided over the proposed construction of a sprint car racetrack in the​ town, as shown in the table on the right. Support
Racetrack
Oppose Racetrack
Live in Town
​2,474
384
Live in Surrounding Area
​3,550
620
If a newspaper reporter randomly selects a person to interview from these​ people,
a. What is the probability that the person supports the​ racetrack? Write your answer as a simplified fraction.
b. What are the odds in favor of the person supporting the​ racetrack?

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

the battle of gettysburg is considered the "high water mark" of the confederacy because it marked the greatest advance of the confederate forces against the union.

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb