Answer:This problem is about linear equations. We assume Dale drive X miles, and the total cost is $Y, then we can get:
Plan I: Y=38+0.11X
Plan II: Y=49+0.07X
When both plans cost the same, 38+0.11X=49+0.07X. We will get X = 275miles, and Y=$68.25
Step-by-step explanation:
Answer:
Step-by-step explanation:
Hello!
So you have a new type of shoe that lasts presumably longer than the ones that are on the market. So your study variable is:
X: "Lifetime of one shoe pair of the new model"
Applying CLT:
X[bar]≈N(μ;σ²/n)
Known values:
n= 30 shoe pairs
x[bar]: 17 months
S= 5.5 months
Since you have to prove whether the new shoes last more or less than the old ones your statistical hypothesis are:
H₀:μ=15
H₁:μ≠15
The significance level for the test is given: α: 0.05
Your critical region will be two-tailed:


So you'll reject the null Hypothesis if your calculated value is ≤-1.96 or if it is ≥1.96
Now you calculate your observed Z-value
Z=<u>x[bar]-μ</u> ⇒ Z=<u> 17-15 </u> = 1.99
σ/√n 5.5/√30
Since this value is greater than the right critical value, i.e. Zobs(1.99)>1.96 you reject the null Hypothesis. So the average durability of the new shoe model is different than 15 months.
I hope you have a SUPER day!
Answer:
The fact that the expressions are equivalent means that for any value of the variables, the two expressions have the same solution.
Step-by-step explanation:
Answer: Man sorry I don’t know how to solve that go to a math solver or something
Step-by-step explanation:
Answer:
24 miles UwU
Step-by-step explanation: G'day mate!