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3 years ago

What the answer for $9; 10% increase

Mathematics

2 answers:

NeTakaya3 years ago

8 0

48 cookies; 25 increase

harkovskaia [24]3 years ago

7 0

10% = 0.10
To get 10% of 9, you multiply 9 times 0.1 Which equals 0.9 then add the original $9 to get a final answer of $9.9

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23. Find the value of x.<br> (5x + 4)' X (8x - 71)

r-ruslan [8.4K]

Answer:

x = 71/8

Step-by-step explanation:

Solve for x:

5 X (8 x - 71) = 0

Divide both sides by 5 X:

8 x - 71 = 0

Add 71 to both sides:

8 x = 71

Divide both sides by 8:

Answer: x = 71/8

8 0

3 years ago

Can someone please help 4x + 2x + 1 = 2x - x + 21

erica [24]

Answer: Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

4*x+2*x+1-(2*x-x+21)=0

Step by step solution :

STEP

Pulling out like terms

1.1 Pull out like factors :

5x - 20 = 5 • (x - 4)

Equation at the end of step

STEP

Equations which are never true:

2.1 Solve : 5 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

2.2 Solve : x-4 = 0

Add 4 to both sides of the equation :

x = 4

One solution was found :

x = 4

3 0

3 years ago

To solve by completing the square, what needs to be moved in this equation?

Elza [17]

Answer:

To complete the square, we need to have the equation set up like this:

x^2 +4x =9

To do that, we'd need to move the 4x to the left side producing

4x + x^2 = 9

And that can be rewritten as

x^2 +4x =9

So the term that needs to be moved is -4x

https://www.1728.org/quadr2.htm

Step-by-step explanation:

7 0

3 years ago

Consider a discrete random variable x with pmf px (1) = c 3 ; px (2) = c 6 ; px (5) = c 3 and 0 otherwise, where c is a positive

PtichkaEL [24]

Looks like the PMF is supposed to be

$\mathbb P(X=x)=\begin{cases}\dfrac c3&\text{for }x\in\{1,5\}\\\\\dfrac c6&\text{for }x=2\\\\0&\text{otherwise}\end{cases}$

which is kinda weird, but it's not entirely clear what you meant...

Anyway, assuming the PMF above, for this to be a valid PMF, we need the probabilities of all events to sum to 1:

$\displaystyle\sum_{x\in\{1,2,5\}}\mathbb P(X=x)=\dfrac c3+\dfrac c6+\dfrac c3=\dfrac{5c}6=1\implies c=\dfrac65$

Next,

$\mathbb P(X>2)=\mathbb P(X=5)=\dfrac c3=\dfrac25$

$\mathbb E(X)=\displaystyle\sum_{x\in\{1,2,5\}}x\,\mathbb P(X=x)=\dfrac c3+\dfrac{2c}6+\dfrac{5c}3=\dfrac{7c}3=\dfrac{14}5$

$\mathbb V(X)=\mathbb E\bigg((X-\mathbb E(X))^2\bigg)=\mathbb E(X^2)-\mathbb E(X)^2$
$\mathbb E(X^2)=\displaystyle\sum_{x\in\{1,2,5\}}x^2\,\mathbb P(X=x)=\dfrac c3+\dfrac{4c}6+\dfrac{25c}3=\dfrac{28c}3=\dfrac{56}5$
$\implies\mathbb V(X)=\dfrac{56}5-\left(\dfrac{14}5\right)^2=\dfrac{84}{25}$

If $Y=X^2+1$ , then $X^2=Y-1\implies X=\sqrt{Y-1}$ , where we take the positive root because we know $X$ can only take on positive values, namely 1, 2, and 5. Correspondingly, we know that $Y$ can take on the values $1^2+1=2$ , $2^2+1=5$ , and $5^2+1=26$ . At these values of $Y$ , we would have the same probability as we did for the respective value of $X$ . That is,

$\mathbb P(Y=y)=\begin{cases}\dfrac c3&\text{for }y=2\\\\\dfrac c6&\text{for }y=5\\\\\dfrac c3&\text{for }y=26\\\\0&\text{otherwise}\end{cases}$

Part (5) is incomplete, so I'll stop here.

8 0

3 years ago

What does endpoint mean

emmasim [6.3K]

Where you stop/finish. hope that helps :)

6 0

3 years ago

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