1.find the sum. (3^3+5x^2+3x-7)+(8x-6x^2+6) 2. Find the Difference: (8x-4x^2+3\)-(x^3+7x^2+3x-8)

1 answer:

3 0

$(3^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=27\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=-x^2+11x+26$

or if you mean (3x^3+5x^2+3x-7)+(8x-6x^2+6):

$(3x^3+5x^2+3x-7)+(8x-6x^2+6)=\\\\=3x^3\ \underline{+\,5x^2}\ \underline{ \underline{+\,3x}}-7\ \underline{\underline{\,+8x}}\ \underline{-\,6x^2}+6=\\\\=3x^3-x^2+11x-1$

$(8x-4x^2+3)-(x^3+7x^2+3x-8)=\\\\=\underline{8x}\ \underline{\underline{-\ 4x^2}}+3-x^3\ \underline{\underline{-\,7x^2}}\ \underline{-\,3x}+8=\\\\=-x^3-11x^2+5x+10$

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Answer:

b) H0:μ=90, H1:μ≠90

Step-by-step explanation:

The corporation wants to know if the mean number of days is different from the 90 days claimed.

This means that at the null hypothesis, we test if the mean number is the claimed value of 90, and so:

$H_0: \mu = 90$

The corporations wants to know if the mean number is different, and thus, at the alternate hypothesis, we test if the mean number is different from 90, that is:

$H_1: \mu \neq 90$