Evaluate the function rule for the given value f(x)=15x for x=3
2 answers:
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The zero product property tells us that if the product of two or more factors is zero, then each one of these factors CAN be zero.
For more context let's look at the first equation in the problem that we can apply this to:
Through zero property we know that the factor
can be equal to zero as well as 
. This is because, even if only one of them is zero, the product will immediately be zero.
The zero product property is best applied to factorable quadratic equations in this case.
Another factorable equation would be
since we can factor out 
and end up with 
. Now we'll end up with two factors, and 
, which we can apply the zero product property to.
The rest of the options are not factorable thus the zero product property won't apply to them.
For more context let's look at the first equation in the problem that we can apply this to:
Through zero property we know that the factor
The zero product property is best applied to factorable quadratic equations in this case.
Another factorable equation would be
The rest of the options are not factorable thus the zero product property won't apply to them.
Answer:
<em>Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom</em>
<em>The null hypothesis is accepted . </em>
<em>Assume the population variances are approximately the same</em>
<u><em>Step-by-step explanation:</em></u>
<u>Explanation</u>:-
Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds
<em>The first sample size 'n₁'= 20</em>
<em>mean of the first sample 'x₁⁻'= 17.53 pounds</em>
<em>standard deviation of first sample S₁ = 3.2 pounds</em>
Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds
<em>The second sample size n₂ = 24</em>
<em>mean of the second sample "x₂⁻"= 14.89 pounds</em>
<em>standard deviation of second sample S₂ = 2.7 poun</em>ds
<u><em>Null hypothesis</em></u><u>:-</u><u><em>H₀</em></u><em>: The Population Variance are approximately same</em>
<u><em>Alternatively hypothesis</em></u><em>: H₁:The Population Variance are approximately same</em>
<em>Level of significance ∝ =0.05</em>
<em>Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42</em>
<em>Test statistic :-</em>
<em> </em>
<em> where </em>
<em> substitute values and we get S² = 40.988</em>
<em> </em><em></em>
<em> </em> t = 1.3622
Calculated value t = 1.3622
Tabulated value 't' = 2.081
Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom
<u><em>Conclusion</em></u>:-
<em>The null hypothesis is accepted </em>
<em>Assume the population variances are approximately the same.</em>
<em> </em>
<em> </em>
<em> </em>