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).
8.2c−0.9+(5.4−18d+7.2c)÷0.06 Distribute:<span> =<span><span><span><span><span><span><span>8.2c</span>+</span>−0.9</span>+<span>120c</span></span>+</span>−<span>300d</span></span>+<span>90
</span></span></span> <span>=<span><span><span>(<span><span>8.2c</span>+<span>120c</span></span>)</span>+<span>(<span>−<span>300d</span></span>)</span></span>+<span>(<span><span>−0.9</span>+90</span>)</span></span></span><span> =<span><span><span>128.2c</span>+<span>−<span>300d</span></span></span>+89.1</span></span> Answer:<span>=<span><span><span>128.2c</span>−<span>300d</span></span>+<span>89.1</span></span></span>
70 people are in the band
10 flute players
20 trombone players
10 drummers
30 trumpet players
376 because 276 + 100 is 376
