Step-by-step explanation:
just look at the graphic : what value do we have on the cost side, when we have 1 pounds on the amount side ?
all the answer options specify a price per pound. that is why we only need to focus on the price for one pound.
we see in the graphic, that for 1 pound the price is $4.
just to make sure, we can also work with the directly specified point (0.5, 2), which means that 0.5 pounds cost $2.
and the function is clearly a straight line.
so, every point on the function has the same y/x ratio (slope).
therefore,
2/0.5 = 4
now, what point with x=1 has the same ratio ?
y/1 = 4
y = $4
perfect, the results are identical. so, $4 per pound is the correct answer.
Hey there! :)
2(3/5n + 3) - (2/3n - 1)
Simplify.
(2 * 3/5n) + (2 * 3) - (2/3n -1)
Simplify.
6/5n + 6 - 2/3n + 1
Add like terms.
(6/5n - 2/3n) + (6 + 1)
Make the fractions have the same denominator by multiplying fraction 1 (6/5n) by 3/3 and fraction 2 (-2/3n) by 5/5.
6/5n * 3/3 = 18/15n
-2/3n * 5/5 = -10/15n
So, our new equation looks like this : (18/15n - 10/15n) + (6 + 1)
Simplify.
8/15n + 6 —> final answer.
Hope this helped! :)
Answer:
Step-by-step explanation:
Hello!
The commuter is interested in testing if the arrival time showed in the phone app is the same, or similar to the arrival time in real life.
For this, she piked 24 random times for 6 weeks and measured the difference between the actual arrival time and the app estimated time.
The established variable has a normal distribution with a standard deviation of σ= 2 min.
From the taken sample an average time difference of X[bar]= 0.77 was obtained.
If the app is correct, the true mean should be around cero, symbolically: μ=0
a. The hypotheses are:
H₀:μ=0
H₁:μ≠0
b. This test is a one-sample test for the population mean. To be able to do it you need the study variable to be at least normal. It is informed in the test that the population is normal, so the variable "difference between actual arrival time and estimated arrival time" has a normal distribution and the population variance is known, so you can conduct the test using the standard normal distribution.
c.
![Z_{H_0}= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }](https://tex.z-dn.net/?f=Z_%7BH_0%7D%3D%20%5Cfrac%7BX%5Bbar%5D-Mu%7D%7B%5Cfrac%7BSigma%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D)

d. This hypothesis test is two-tailed and so is the p-value.
p-value: P(Z≤-1.89)+P(Z≥1.89)= P(Z≤-1.89)+(1 - P(Z≤1.89))= 0.029 + (1 - 0.971)= 0.058
e. 90% CI

X[bar] ± 
0.77 ± 1.645 * 
[0.098;1.442]
I hope this helps!
The first graph shows that equation.