A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only .9 hours. What is the plane's speed in still air, and how fast is the wind blowing?

Answers

Answer 1

Answer:

Answer:

90 mi/h

Step-by-step explanation:

Calculate speed, distance, or time using the formula d = st, distance equals speed times time. The Speed Distance Time Calculator can solve for the unknown sdt value given two known values.

Time can be entered or solved for in units of seconds (s), minutes (min), hours (hr), or hours and minutes and seconds (hh:mm:ss).

To solve for distance use the formula for distance d = st, or distance equals speed times time.

distance = speed x time

Rate and speed are similar since they both represent some distance per unit time like miles per hour or kilometers per hour. If rate r is the same as speed s, r = s = d/t. You can use the equivalent formula d = rt which means distance equals rate times time.

distance = rate x time

To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time.

speed = distance/time

To solve for time use the formula for time, t = d/s which means time equals distance divided by speed.

time = distance/speed

Plane speed in still air: 120mph

Wind speed: 30pmh

Speed of plane to Bismarck: 90mph

Speed of plane to Fargo: 150mph

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Answers

Answer:

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Describe the transformation from the graph of f(x) = x + 3 to the graph of g(x) = x − 7.

Answers

f(x) shifts left 3
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A high school student took two college entrance exams, scoring 1070 on the SAT and 25 on the ACT. Suppose that SAT scores have a mean of 950 and a standard deviation of 155 while the ACT scores have a mean of 22 and a standard deviation of 4. Assuming the performance on both tests follows a normal distribution, determine which test the student did better on.

Answers

Answer:

Due to the higher z-score, he did better on the SAT.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

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

$Z = (X - \mu)/(\sigma)$

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 pvalue, we get the probability that the value of the measure is greater than X.

Determine which test the student did better on.

He did better on whichever test he had the higher z-score.

SAT:

Scored 1070, so $X = 1070$