F<span> + </span>g)(x<span>) = </span>f(x<span>) + </span>g(x); (f<span> - </span>g)(x<span>) = </span>f(x<span>) - </span>g(x): (f<span> · </span>g)(x<span>) = </span>f(x<span>) · </span>g(x<span>) ..., let </span>f(x) = 5x+2<span> and </span>g(x<span>) = </span><span>x2</span>-1. <span>4. </span>f(4)=5(4)+2<span>=22 and </span>g(4)=42-1=15 ... (fog)(x<span>) = </span>f<span> [ </span>g(x<span>) ] = </span>f<span> [ 4x2 ] = sqrt( </span><span>4.2</span><span> ) = </span>2<span> | </span>x<span> |; (</span>gof)(x<span>) = </span>g<span> [</span>f(x<span>) ] = </span>g [ s
Answer:
A. The distribution of sample means of the differences will be approximately normal if there are at least 30 years of data in the sample and/or if the population of differences in winning times for all years is normal.
Step-by-step explanation:
In other to perform a valid paired test, one of the conditions required is that, data for both groups must be approximately normal. To attain normality, the population distribution for the groups must be normal or based on the central limit theorem, the sample size must be large enough, usually n > 30. Hence, once either of the two conditions are met, the paired sample will be valid.
Subtract 3 first so it would be 7x=15
then divide by 7 so x would be approximately 2.14
Answer:
473 rounded to the nearest hundredth, would be 400.
