You and two friends pay $40 for tickets. The cost was divided 3 ways in the ratio 1 : 3 : 6. How much did each person pay?
1 answer:
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Answer:
Step-by-step explanation: <u>Speed</u> is distance an object travelled per unit time.
It is represented by letter v and can assume various units.
1)
v = 18 miles per hour
Her average speed is 18 mph.
2) Rearraging formula for distance:
d = v.t
d = 70 * 4
d = 280 miles
You would travel 280 miles.
3)
v = 52.5 mph
My average speed is 52.5 mph.
4) Rearranging formula to determine time:
t = 2.34 hours
It will take 2 hours and half to arrive in Leeds. So, we will be expected to arrive at 12:04.
5) The unit change but the way of calculating time is the same.
t = 500 s
or t = 8.34 minutes
It will take, for an athlete to run 2000m, 8.34 minutes.
6) a) All units must match, so for a speed in mph:
45 minutes is 0.75 hours. Total time is 8.75 hours.
v = 560 mph
The average speed is 560 mph.
b)
t = 6.67 hours
Sam will take 6.67 hours to covered the distance.
Answer:
The value of the test statistic is z = 1.39.
The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.
Step-by-step explanation:
Using traditional methods it takes 92 hours to receive an advanced flying license.
This means that at the null hypothesis, it is tested if the mean is of 92, that is:
Test if there is evidence that the technique lengthens the training time
At the alternative hypothesis, it is tested if the mean is more than 92, that is:
The test statistic is:
In which X is the sample mean, is the value tested at the null hypothesis,
is the standard deviation and n is the size of the sample.
92 is tested at the null hypothesis:
This means that
A researcher used the technique on 70 students and observed that they had a mean of 93 hours. Assume the population variance is known to be 36.
This means that
Value of the test statistic:
The value of the test statistic is z = 1.39.
P-value of the test and decision:
The p-value of the test is the probability of finding a sample mean above 93, which is 1 subtracted by the p-value of z = 1.39.
Looking at the z-table, z = 1.39 has a p-value of 0.9177.
1 - 0.9177 = 0.0823.
The p-value of the test is 0.0823 > 0.05, which means that there is not evidence at the 0.05 level that the technique lengthens the training time.