Help Please!! On the 1st and 2nd one. just the decent and independent variable and the equation please!

A student playing a board game believes that a standard six sided die is incorrectly weighted. The student thinks that the numbe

nexus9112 [7]

Answer:

Hₐ = P ( 1 ) > 1/6

Step-by-step explanation:

This is the answer because the question is asking you for the "alternative hypothesis" and that means that it's basically asking for the statement that agrees with what is being said.

Since we know that a student believes that he's seeing the number "1" too often, then we must find the equation that agrees with that statement, which is Hₐ = P ( 1 ) > 1/6.

P.S - I took the test and got a 100%.

6 0

3 years ago

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If f(x)= 2(x-3) -0.2x what is the value of f(-10) PLEASEEE ANSWER FAST

Rashid [163]

Answer:

-6

Step-by-step explanation:

2(-13) - (0.2*(-10))

-26 + 20

= -6

3 0

3 years ago

Identify each function as linear or non-linear need help with both

Illusion [34]

Non linear and linear

8 0

3 years ago

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Mary’s number is eleven more than Jerry’s. The sum of their numbers is 89. What are their numbers?

stich3 [128]

Take 89 minus 11 to find twice of Jerry's number . To find Jerry's number divide the answer by two which is 89-11 =78. 78 divide by 2 is 39 . To find Mary's number add 39 and 11 .
Hence Mary has 50
Jerry has 39

8 0

4 years ago

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How do you integrate xe^(2x)dx?

vesna_86 [32]

You integrate it by parts:

$\large\begin{array}{l} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx\qquad\quad(i)}\\\\\\ \begin{array}{lcl} \mathsf{u=x}&~\Rightarrow~&\mathsf{du=dx}\\\\ \mathsf{dv=e^{2x}\,dx}&~\Leftarrow~&\mathsf{v=\dfrac{1}{2}\,e^{2x}} \end{array} \end{array}$

$\large\begin{array}{l} \mathsf{\displaystyle\int\!u\,dv=u\cdot v-\int\!v\,du}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=x\cdot \frac{1}{2}\,e^{2x}-\int\!\frac{1}{2}\,e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\int\!e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\int\!\dfrac{1}{2}\cdot 2e^{2x}\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{2}\cdot \frac{1}{2}\int\!e^{2x}\cdot 2\,dx} \end{array}$

$\large\begin{array}{l} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\int\!e^{2x}\cdot 2\,dx}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\int\!e^w\,dw\qquad(w=2x)}\\\\ \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\,e^w+C}\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathsf{\displaystyle\int\!x\,e^{2x}\,dx=\frac{1}{2}\,x\,e^{2x}-\frac{1}{4}\,e^{2x}+C} \end{array}}\qquad\checkmark \end{array}$

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$\large\textsf{I hope it helps. :-)}$

Tags: integrate indefinite integral product by parts polynomial exponential composite substitution integral calculus

3 0

4 years ago