Answer:
f(n) = f(n − 1) ⋅ 0.6 + 20, n > 0
Step-by-step explanation:
Each week 40% of the olive oil bottles were sold. This means, each week 60% of the bottles were left. In addition to these, 20 new bottles arrived in shipments.
So, every week the number of shipments was:
60% of the shipments of previous week + 20
In equation form this can written as:
Shipments in a week = 60% of shipments of previous week + 20
For, nth week we can re-write this equation as:
f(n) = 60% of f(n - 1) + 20
or
f(n) = f(n - 1) ⋅ 0.60 + 20
For example for first week, replace n by 1 to find the number of shipments in week 1 and same goes for next weeks.
Answer:
10 terms
Step-by-step explanation:
equate the sum formula to 55 and solve for n
n(n + 1) = 55 ( multiply both sides by 2 to clear the fraction )
n(n + 1) = 110 ← distribute parenthesis on left side
n² + n = 110 ( subtract 110 from both sides )
n² + n - 110 = 0 ← in standard form
Consider the factors of the constant term (- 110) which sum to give the coefficient of the n- term (+ 1)
the factors are + 11 and - 10 , since
11 × - 10 = - 110 and 11 - 10 = + 1 , then
(n + 11)(n - 10) = 0 ← in factored form
equate each factor to zero and solve for n
n + 11 = 0 ⇒ n = - 11
n - 10 = 0 ⇒ n = 10
However, n > 0 , then n = 10
number of terms which sum to 55 is 10
The correct answer would be a Right triangle
Answer:
b140
Step-by-step explanation:
14*10=140 that's all
Answer: (C) 0.1591
Step-by-step explanation:
Given : A manufacturer of radial tires for automobiles has extensive data to support the fact that the lifetime of their tires follows a normal distribution with
![\mu=42,100\text{ miles}](https://tex.z-dn.net/?f=%5Cmu%3D42%2C100%5Ctext%7B%20miles%7D)
![\sigma=2,510\text{ miles}](https://tex.z-dn.net/?f=%5Csigma%3D2%2C510%5Ctext%7B%20miles%7D)
Let x be the random variable that represents the lifetime of the tires .
z-score : ![z=\dfrac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
For x= 44,500 miles
![z=\dfrac{44500-42100}{2510}\approx0.96](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B44500-42100%7D%7B2510%7D%5Capprox0.96)
For x= 48,000 miles
![z=\dfrac{48000-42100}{2510}\approx2.35](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7B48000-42100%7D%7B2510%7D%5Capprox2.35)
Using the standard normal distribution table , we have
The p-value : ![P(44500](https://tex.z-dn.net/?f=P%2844500%3Cx%3C48000%29%3DP%280.96%3Cz%3C2.35%29)
![P(z](https://tex.z-dn.net/?f=P%28z%3C2.35%29-P%28z%3C0.96%29%3D%200.9906132-0.8314724%3D0.1591408%5Capprox0.1591)
Hence, the probability that a randomly selected tire will have a lifetime of between 44,500 miles and 48,000 miles = 0.1591