Answer:
Annike's bike trip started at 7:00, if she and Celia both start at the same time you would multiply Annike's average speed by Celia's average speed to find where they would have completed the same number of miles
equation: 11×14=154 14×11=154
Step-by-step explanation:
counting by 11:
11 22 33 44 55 66 77 88 99 110 121 132 143 <em><u>154</u></em>
11 × 14 = 154
counting by 14:
14 28 42 56 70 84 98 112 126 140 <em><u>154</u></em>
14 × 11 = 154
The probability of you running into a Mankey Pokemon and then a Pidgey Pokemonis 4/27.
A customer borrowed $2000 and then a further $1000 both repayble in 12 months. What would he have saved if he had taken out one loan for $3000 repayable in 12 months?
He took two different loans, it charged him loan processing fee twice, two-time documentation process, and of course, extra time spent for second loan. Instead, he could take single loan of $3000 with one-time processing fee, one-time documentation process, and time-saving also.
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!