PLEASE HELP .

ωнι¢н σƒ тнє ƒσℓℓσωιηg ιѕ тнє ρσѕιтιση ѕ(10) σƒ тнє мσνιηg вσ∂у αт тιмє т=10ѕ, gινєη тнαт ιтѕ α¢¢єℓєяαтιση, α(т)=9.8 м/ѕ^2 , ιηι

jek_recluse [69]

I am eng k ...................

8 0

3 years ago

A total of 300 pine and maple trees will be planted in a park. There will be 2 pine trees planted for every 3 maple trees plante

mars1129 [50]

Answer:

120

Step-by-step explanation:

2 of every 5 trees are pine trees.

2/5 · 300 = 120

120 pine trees are going to be planted in the park.

5 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

4 years ago

In a certain geographic area a heavy rainfall occurs on the average two times per three months. Find the probability that during

SVETLANKA909090 [29]

This is a Poisson Distribution
Average over 3 months = 2
Average over 1 month = 2/3
Average over 2 months = 4/3 = λ
More than four heavy rainfalls means 5, 6, 7,…
P(5) = (e^-λ x λ^5)/5! = 0.009257
P(6) = 0.002057
P(7) = 0.000392
P(8) = 0.000065
P(9) = 0.000010
P(10) = 0.000001
P(more than 4 heavy storms) = 0.01178 (4sf)

8 0

2 years ago

5, for a = 5 and b = 6.

andrezito [222]

im neidd so 4567Answer:

Step-by-step explanation:

7 0

3 years ago