Answer:
Q5) 7
Q6) 4
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Step-by-step explanation:
ANSWER

EXPLANATION
The given rational function is

We need to rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of

which is

When we rationalize we obtain:

The denominator is now a difference of two squares:

We apply this property to get


This simplifies to

Or

Answer:
Option C.
Step-by-step explanation:
we know that
The compound interest formula is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
substitute in the formula above
Answer:
Yes. The data provide enough evidence to support the claim that the mean weight of one-year-old boys is greater than 25 pounds.
P-value=P(t>2.84)=0.0024
Step-by-step explanation:
Hypothesis test on the population mean.
The claim is that the mean weight of one-year-old boys is greater than 25 pounds.
Then, the null and alternative hypothesis are:

The significance level is α=0.05.
The sample size is n=354. The sample mean is 25.8 pounds and the sample standard deviation is 5.3 pounds. As the population standard deviation is estimated from the sample standard deviation, we will use a t-statistic.
The degrees of freedom are:

The t-statistic is:

For a right tailed test and 353 degrees of freedom, the P-value is:

As the P-value is smaller than the significance level, the effect is significant and the null hypothesis is rejected.
There is enough evidence to support the claim that the mean weight of one-year-old boys is greater than 25 pounds.
Answer:
(5x+2)(3x-2)
Step-by-step explanation: