$D:x+2\neq0\to x\neq-2\ and\ x-3\neq0\to x\neq3\\\\\frac{3}{x+2}+\frac{4}{x-3}=\frac{3(x-3)}{(x+2)(x-3)}+\frac{4(x+2)}{(x+2)(x-3)}=\frac{3(x)+3(-3)+4(x)+4(2)}{(x)(x)+(x)(-3)+(2)(x)+2(-3)}\\\\=\frac{3x-9+4x+8}{x^2-3x+2x-6}=\boxed{\frac{7x-1}{x^2-x-6}}$

3 0

4 years ago

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A newsletter publisher believes that 30% of their readers own a Rolls Royce. A testing firm believes this is inaccurate and perf

fredd [130]

Answer:

The evidence is not sufficient at the 0.02 level of significance to reject the claim that the percentage is 30%.

Step-by-step explanation:

Let's first state the null and alternative hypotheses for this research;

H0: p = 0.30

Ha: p ≠ 0.30

Now, since the bull hypothesis is the claim and he fails to reject the null hypothesis from the research, we will conclude that;

The evidence is not sufficient at the 0.02 level of significance to reject the claim that the percentage is 30%.

5 0

3 years ago

HEEELLLLP GRADES ARE DUE In 4 HOURS 0.13(x + 2000 ) = 6,240 What is x?

Firdavs [7]

Answer:

4240 is what x is eq to

Step-by-step explanation:

6 0

3 years ago