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devlian [24]
2 years ago
9

What is the variable in 3.2x + 5.1x = 41.5

Mathematics
1 answer:
Alex2 years ago
7 0
3.2x + 5.1x = 41.5
- add 3.2x and 5.1x
8.3x = 41.5
- divide both sides by 8.3
x = 5
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12. If you flip a dime 50 times:
nata0808 [166]

Answer: 25 heads 25 tails

Step-by-step explanation: you would have a 50/50 chance to get heads or tails so sence there are 2 sides you would get 25 heads and 25 tails

4 0
3 years ago
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The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p
Juliette [100K]

Check the picture below.


based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,


\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)


\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}


now, we can plug those values on A = (1/2)bh,


\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5

7 0
3 years ago
PLEASE HELP LIKE RN, thank you so much
Komok [63]

Answer:

Facts.

Step-by-step explanation:

4 0
3 years ago
Use the method of completing the square to transform the quadratic equation into the equation form (x + p)2 = q. 3 + x - 3x2 = 9
Triss [41]
To <span>transform the quadratic equation into the equation form (x + p)2 = q we shall proceed as follows:
3+x-3x^2=9
putting like terms together we have:
-3x^2+x=6
dividing through by -3 we get:
x^2-x/3=-2
but
c=(b/2a)^2
c=(-1/6)^2=1/36
thus the expression will be:
x^2-x/3+1/36=-2+1/36
1/36(6x-1)</span>²=-71/36

the answer is:
1/36(6x-1)²=-71/36
7 0
3 years ago
Suppose a basketball player has made 231 out of 361 free throws. If the player makes the next 2 free throws, I will pay you $31.
statuscvo [17]

Answer:

The expected value of the proposition is of -0.29.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either the player will make it, or he will miss it. The probability of a player making a free throw is independent of any other free throw, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Suppose a basketball player has made 231 out of 361 free throws.

This means that p = \frac{231}{361} = 0.6399

Probability of the player making the next 2 free throws:

This is P(X = 2) when n = 2. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{2,2}.(0.6399)^{2}.(0.3601)^{0} = 0.4095

Find the expected value of the proposition:

0.4095 probability of you paying $31(losing $31), which is when the player makes the next 2 free throws.

1 - 0.4059 = 0.5905 probability of you being paid $21(earning $21), which is when the player does not make the next 2 free throws. So

E = -31*0.4095 + 21*0.5905 = -0.29

The expected value of the proposition is of -0.29.

3 0
2 years ago
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