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:
A
x=0.625y
y=1.6x
B
y=3.2
Step-by-step explanation:
Answer: each girl received 30 stickers.
Step-by-step explanation:
Let x represent the number of stickers that the boys received.
Let y represent the number of stickers that the girls received.
The total number of children was 42. 2/3 of the children were boys. It means that the number of boys were
2/3 × 42 = 28
The number if girls would be
42 - 28 = 14
The total number of stickers that the school principal shared was 840. It means that
28x + 14y = 840- - - - - - - - -1
Each of the boys received the same number of stickers while each girl received twice as many as each boy. It means that
y = 2x
Substituting y = 2x into equation 1, it becomes
28x + 14 × 2x = 840
28x + 28x = 840
56x = 840
x = 840/56
x = 15
y = 2x = 15 × 2
y = 30