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grigory [225]
3 years ago
13

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

1

:

Pulling out like terms

1.1 Pull out like factors :

5x - 20 = 5 • (x - 4)

Equation at the end of step

1

:

STEP

2

:

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
Read 2 more answers
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