Helppp meeee pleaseeee

Mathematics

1 answer:

finlep [7]2 years ago

8 0

Answer:

0.3907 I think hope this helps

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Simplify the expression -7(-10-6)+1v+10v

ollegr [7]

Answer:

+112 +11v

Step-by-step explanation:

-7(-10-6)+1v+10v =

(-10-6) = -16

-7(-16) +1v +10v =

+112 +1v +10v =

+112 +11v

8 0

3 years ago

The owner of a local nightclub has recently surveyed a random sample of n = 250 customers of the club. She would now like to det

seraphim [82]

Answer:

The required hypothesis to test is $H_0: \mu\leq 30$ and $H_1: \mu>30$ .

Step-by-step explanation:

Consider the provided information.

It is given that the nightclub has recently surveyed a random sample of n = 250 customers of the club.

She would now like to determine whether or not the mean age of her customers is over 30.

Null hypotheses is represents as $H_0$ . Thus the null hypotheses is shown as:

$H_0: \mu\leq 30$

The alternative hypotheses is represents as $H_1$ . Thus the alternative hypotheses is shown as:

$H_1: \mu>30$

Hence, the required hypothesis to test is $H_0: \mu\leq 30$ and $H_1: \mu>30$ .

8 0

3 years ago

Y=(2x-3)^5(2-x^4)^3 differentiate the function please

BARSIC [14]

$\bf y=(2x-3)^5(2-x^4)^3\\\\
-------------------------------\\\\
\cfrac{dy}{dx}=\stackrel{\textit{product rule}}{[5(2x-3)^4\cdot 2(2-x^4)3]~~+~~[(2x-3)^5[3(2-x^4)^2(-4x^3)]]}
\\\\\\
\cfrac{dy}{dx}=10(2x-3)^4(2-x^4)^3~~-~~12x^3(2x-3)^5(2-x^4)^2
\\\\\\
\cfrac{dy}{dx}=\stackrel{\textit{common factor}}{2(2x-3)^4(2-x^4)^2}~[5(2-x^4)-6x^3(2x-3)]
\\\\\\
\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2~[10-5x^4-12x^4+18x^3]
\\\\\\
\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2(10-17x^4+18x^3)$