To answer the question presented above, the conversion factor for km to feet is 1 kilometers is equal to 3280.84 feet. From the given,
(12 km) x (3280.84 ft / 1 km) = 39370.08 feet
Thus, Joe traveled an approximate distance of 39370 feet.
Answer:
Step-by-step explanation:
In the normal distribution curve, the mean is in the middle and each line to the left and to the right of that mean represent 1- and 1+ the standard deviation. If our mean is 400, then 400 + 50 = 450; 450 + 50 = 500; 500 + 50 = 550. Going from the mean to the left, we subtract the standard deviation and 400 - 50 = 350; 350 - 50 = 300; 300 - 50 = 250. We are interested in the range that falls between 350 and 450 as a percentage. That range represents the two middle sections, each containing 34% of the data. So the total percentage of response times is 68%. We are looking then for 68% of the 144 emergency response times in town. .68(144) = 97.92 or 98 emergencies that have response times of between 350 and 450 seconds.
Answer:
(E) None of these above are true.
Step-by-step explanation:
Married = 74% or 0.74
College graduates = 42% or 0.42
pr(married | college graduates) = 0.56
(A) These events are pairwise disjoint. This is false. Pairwise disjoint are also known as mutually exclusive events. Here we can see that both events are occurring at same time.
(B) These events are independent events. This is also false.
(C) These events are both independent and pairwise disjoint. False
(D) A worker is either married or a college graduate always. False
Here Probability(A or B) shall be 1
= Pr(A) + Pr(B) - Pr( A and B) = 0.74 + 0.42 - 0.56 * 0.42 = 0.9248
This is not equal to 1.
(E) None of these above are true. This is true.
After an hour (4pm to 5pm) the temperature drops 18e⁻⁰˙⁶=9.9℉, so the temperature after an hour will be 68-9.9=58.1℉.
(The change -18e⁻⁰˙⁶ is negative indicating a drop in temperature.)
Answer:
a. z = 2.00
Step-by-step explanation:
Hello!
The study variable is "Points per game of a high school team"
The hypothesis is that the average score per game is greater than before, so the parameter to test is the population mean (μ)
The hypothesis is:
H₀: μ ≤ 99
H₁: μ > 99
α: 0.01
There is no information about the variable distribution, I'll apply the Central Limit Theorem and approximate the sample mean (X[bar]) to normal since whether you use a Z or t-test, you need your variable to be at least approximately normal. Considering the sample size (n=36) I'd rather use a Z-test than a t-test.
The statistic value under the null hypothesis is:
Z= X[bar] - μ = 101 - 99 = 2
σ/√n 6/√36
I don't have σ, but since this is an approximation I can use the value of S instead.
I hope it helps!
