3 * 5 = 15 miles during weekdays
6.5*2 = 13 miles on weekends
13 +15 = 28 miles total
28 / 7 = 4 miles average per day
Time out = d/so = d/523
time back = d/sb = d/415
t = time out + time back
11 = d/523 + d/415 find common deneminator
11 = (415d + 523d)/(523*415)
11*523*415 = 415d + 523d
2387495 = 938d
d = 2545 miles
hope this help
Answer:
D) 
Step-by-step explanation:
Basically, if you plug in the x and y-coordinates into an equation, both sides should work out to be equal to each other:
Since
, we can eliminate choice A
Since
, we can eliminate choice B
Since
, we can eliminate choice C
Therefore, since
, then D is correct. Also, 
Answer:
Step-by-step explanation:
Hello!
The objective of this experiment is to test if two different foam-expanding agents have the same foam expansion capacity
Sample 1 (aqueous film forming foam)
n₁= 5
X[bar]₁= 4.7
S₁= 0.6
Sample 2 (alcohol-type concentrates )
n₂= 5
X[bar]₂= 6.8
S₂= 0.8
Both variables have a normal distribution and σ₁²= σ₂²= σ²= ?
The statistic to use to make the estimation and the hypothesis test is the t-statistic for independent samples.:
t= ![\frac{(X[bar]_1 - X[bar]_2) - (mu_1 - mu_2)}{Sa*\sqrt{\frac{1}{n_1} + \frac{1}{n_2 } } }](https://tex.z-dn.net/?f=%5Cfrac%7B%28X%5Bbar%5D_1%20-%20X%5Bbar%5D_2%29%20-%20%28mu_1%20-%20mu_2%29%7D%7BSa%2A%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_1%7D%20%2B%20%5Cfrac%7B1%7D%7Bn_2%20%7D%20%7D%20%7D)
a) 95% CI
(X[bar]_1 - X[bar]_2) ±
*
Sa²=
=
= 0.5
Sa= 0.707ç

(4.7-6.9) ± 2.306* 
[-4.78; 0.38]
With a 95% confidence level you expect that the interval [-4.78; 0.38] will contain the population mean of the expansion capacity of both agents.
b.
The hypothesis is:
H₀: μ₁ - μ₂= 0
H₁: μ₁ - μ₂≠ 0
α: 0.05
The interval contains the cero, so the decision is to reject the null hypothesis.
<u>Complete question</u>
a. Find a 95% confidence interval on the difference in mean foam expansion of these two agents.
b. Based on the confidence interval, is there evidence to support the claim that there is no difference in mean foam expansion of these two agents?
Answer:
y>5/6
Step-by-step explanation:
5y+3>-7y+13
5y+7y>13-3
12y/12>10/12
y>5/6
This is an inequality, which shows where the variable y does or does not exist
from this we can say
y: (-infinity,5/6) (5/6, infinity]
or, y does not exist from (-infinity,5/6)
and does exist from (5/6, infinity]