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Natalija [7]
3 years ago
13

It costs 33 tickets to play a game at the fair. If you spent 2121 tickets on the game, which equation could you use to solve for

the number of games, g?
Mathematics
2 answers:
ipn [44]3 years ago
6 0

Answer:

f(g)= 33g-2121

Step-by-step explanation:


Mademuasel [1]3 years ago
3 0

Answer:

3g=21

Step-by-step explanation:

when you isolate g you divide by 3 on both sides giving you the number of games you played

I also got it right on ttm

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Please help asap !!!!!
Natasha2012 [34]

Answer:

<em>XY = 92 units</em>

Step-by-step explanation:

<u>Similar Shapes</u>

Two shapes are similar if all their corresponding side measures are in the same proportion.

The triangles UVW and YVX are similar because their side lengths are in the proportion 1:2, given the tick marks provided in the drawing.

This means that the measure of VX is twice the measure of VW,

The measure of YV is twice the measure of UV

The measure of XY is twice the measure of UW

This last proportion gives the equation:

z + 46 = 2z

Solving for z:

z = 46

Thus, XY = z+46 = 92

XY = 92 units

6 0
3 years ago
Find the slope between the pair of points (1,3) and (7,2)
agasfer [191]

Answer:

             \bold{m=-\dfrac16}

Step-by-step explanation:

\bold{slope\, (m)=\dfrac{change\ in\ Y}{change\ in\ X}=\dfrac{y_2-y_1}{x_2-x_1}}

(1, 3)    ⇒   x₁ = 1,  y₁ = 3

(7, 2)    ⇒   x₂ = 7,  y₂ = 2

So the slope:

                    \bold{m=\dfrac{2-3}{7-1}=\dfrac{-1}6=-\dfrac16}

5 0
3 years ago
Solve for e.<br><br><br>9e + 4 = -5e + 14 + 13e
sladkih [1.3K]

Answer:

10.

Step-by-step explanation:

9e + 4 = -5e + 14 + 13e

9e + 5e - 13e = 14 - 4

e = 10

5 0
2 years ago
Read 2 more answers
PLEASE HELPPP!! I WILL MARK YOU AS BRAINLIEST!
sveticcg [70]

Answer:

this should be right

Step-by-step explanation:

angle a is 60 degrees

4 0
3 years ago
Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen
k0ka [10]

Let a,b,c be the randomly selected lengths. Without loss of generality, suppose a[tex]P(A + B \ge C) = P(A + B - C \ge 0)

where A,B,C are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)

so that the joint density function is

P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}

where [0,1]^3=[0,1]\times[0,1]\times[0,1] is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

a + b - c = 0

with (a,b,c)\in\Bbb R^3. This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region a+b-c\ge0, since 0+0-1=-1. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}

5 0
2 years ago
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