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marishachu [46]
4 years ago
6

Show work and simplify the equation

Mathematics
1 answer:
disa [49]4 years ago
5 0
Hello,

if \ a \neq -3, a \neq 1, a \neq -5\\

 \dfrac{2a^2+5a-3}{a^2+8a+15}  *  \dfrac{a^2+4a-5}{3a^2-a-2} \\


= \dfrac{(a+3)(2a-1)}{(a+5)(a+3)} *  \dfrac{(a+5)(a-1)}{(a-1)(3a+2)} \\

= \dfrac{2a-1}{3a+2}

explains:

1)
a^2+4a-5=a^2+5a-a-5\\
=a(a+5)-(a+5)\\
=(a+5)(a-1)




2)
a^2+8a+15=a^2+5a+3a+15\\
=a(a+5)+3(a+5)\\
=(a+5)(a+3)


3)
a^2+4a-5=a^2+5a-a-5\\
=a(a+5)-(a+5)\\
=(a+5)(a-1)


4)
3a^2-a-2=3a^2-3a+2a-2\\
=3a(a-1)+2(a-1)\\
=(a-1)(3a+2)



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Answer:

\left(g/f\right)\left(x\right)=\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}

Step-by-step explanation:

f(x)=4x+1

g\left(x\right)=x^2\:-\:5

As

(g/f)(x) = g(x) / f(x)

            =\:\frac{x^2\:-\:5}{4x+1}\:\:\:\:\:\:

            =\frac{x}{4}+\frac{-\frac{x}{4}-5}{4x+1}

              =\frac{x}{4}-\frac{1}{16}+\frac{-\frac{79}{16}}{4x+1}

               =\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}

Therefore,

\left(g/f\right)\left(x\right)=\frac{x}{4}-\frac{1}{16}-\frac{79}{16\left(4x+1\right)}

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Answer:

A. Null and alternative hypothesis:

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The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

H_0: \mu_d=0\\\\H_a:\mu_d\neq 0

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

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The degrees of freedom for this sample size are:

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As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

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