Tracy has a cell phone plan that provides 250 free minutes each month for a flat rate of 29$. For any minutes over 250 tracy us
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If period of 
is one-half the period of 
and
<span> has a period of 2π, then 
and 
.
</span>
To find the period of sine function
we use the rule 
.
<span /><span />
f is sine function where f (0)=0, then c=0; with period
, then 
, because 
.
To find a we consider the condition
, from where 
.
If the amplitude of is twice the amplitude of , then has a product factor twice smaller than and while period of <span> </span> is 2π and g(0)=0, we can write 
.
<span>
</span>
To find the period of sine function
<span /><span />
f is sine function where f (0)=0, then c=0; with period
To find a we consider the condition
If the amplitude of
Answer:
Part A: The null hypothesis failed to be rejected.
Part B: The null hypothesis failed to be rejected.
Step-by-step explanation:
We have an hypothesis test with null and alternative hypothesis H0: p = 0.5 versus Ha: p > 0.5, which has the test statistic z=1.15.
Part A: If the significance level is 0.05, the conclusion depends on the P-value.
If the P-value is below 5%, the null hypothesis is rejected.
The P-value for this right-tailed tes and z=1.15 is:
The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.
Part B: In this case, the significance level is 0.01 and, as the alternative hypothesis is defined with an unequal sign, the test is two-tailed.
This changes the way we calculate the P-value, as we need to compute the two tails.
The P-value is:
The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.
