Step-by-step explanation:
A study was to be undertaken to determine if a particular training program would improve physical fitness. A sample of 31 university students was selected to be enrolled in the fitness program. The researchers wished to determine if there was evidence that their sample of students differed from the general population of untrained subjects. The sample mean is 47.4 and a standard deviation of 5.3. The 98% confidence interval is determined and is given as, (45.2, 49.6) .
If the level of confidence is changed to 95%, then the confidence interval will become shorter but the p-value will not change because it is calculated using the test statistic. So the correct answer is (a).
Answer:
use the formula given in the pic and you will get the answer
The simplest way to do this is to set up equivalent fractions. So, you'd do:
=
and then solve for x.
x=52
get a common denominator of 18
5/6 * 3/3 +4/9 * 2/2 - -2*18/18
15/18 +8/18 + 36/18
23/18 + 36/18
59/18
change to a mixed number
18 goes into 59 3 times with 5 left over
3 and 5 /18
He paid him : 24t + 15
for 3 hrs...
24(3) + 15 = 72 + 15 = $ 87
