Please help, I will mark Brainliest.

A pilot flew a 400 mile flight in 2.5 hours flying into the wind. Flying the same rate and with the same wind speed the return t

eduard

Answer: 20 mph

Explanation:

Speed is a physical quantity which is equal to the ratio between the distance covered (d) and the time taken (t):

$v=\frac{d}{t}$

In the first part of the problem, the plane flew a distance of d=400 mi in a time of t=2.5 h. The speed of the plane in this case was the difference between the proper speed of the plane, v, and the speed of the wind, w, since the plane flew opposite to the wind. So we can write:

$v-w=\frac{400mi}{2.5h}=160 mph$ (1)

During the return trip, the plane flew with a speed (v+w), since the wind was on the tail, and it took 2 hours to cover the same distance:

$v+w=\frac{400 mi}{2 h}=200 mph$ (2)

So we have two equations with two unknown variables. From (1), we get

$v=160+w$

Substituting into eq.(2)

$(160+w)+w=200\\160+2w=200\\2w=40\\w=20 mph$

So, the speed of the wind was 20 mph.

3 0

3 years ago

Read 2 more answers

What equals 5015 in multiplacation?

UkoKoshka [18]

To find this out, just simply divide this answer by two.
So, 5015 divided by 2 equals, 2507.5
SO mutiply that by two and you get your needed answer of 5015
~Hope this helped :)

5 0

3 years ago

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The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .