Answer:
The answer is below
Step-by-step explanation:
The question is not complete, but I would show you how to solve it.
Solution:
A circle is the locus of a point such that its distance from a fixed point which is its center is always constant.
The tire has the shape of a circle. Therefore the distance the tire covers if it is pushed around once is the same as the circumference of the tire. The circumference is given by:
Circumference = 2πr; where r is the radius of the tire
Let us assume that the tire has a radius of 7 cm. Hence:
Circumference = 2π(7) = 44 cm
If the tire moves around 5 times, the distance covered = 5 * circumference of tire = 5 * 44 cm = 220 cm
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:
B = 3
Step-by-step explanation:
3x + 15 = 4x + 12
= 15-12=4x-3x
= 3 = x
X = 3
Answer:
140
Step-by-step explanation: