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Jobisdone [24]
3 years ago
8

How do you solve this linear equality?

Mathematics
1 answer:
timama [110]3 years ago
8 0

Answer:

make them into the same fraction problem solve

Step-by-step on:

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You and two friends visit a pizza shop. There are 15 toppings. 5 toppings are meat and the rest are vegetable. Each of you order
andrew-mc [135]
We know that
total toppings=15
<span>meat toppings=5
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<span>The first person has a
5/15 chance of getting a meat topping.

The second person has a
4/14 chance because there are no repeats,

The third person would have a
3/13 chance.
Since there are three multiply
5/15 * 4/14 * 3/13.=60/2730
</span>60/2730-------> divide by 30 both members------> 2/91

the answer is
2/91
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3 years ago
Rewrite equation in slope intercept form 3x+5y=12
pochemuha

Answer:

y= -3/5x + 12/5

Step-by-step explanation:

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3 years ago
Stretch your thinking, solve the story problem. Draw comparison bars. There were 11 hikers here yesterday. There are 4 hikers he
Kryger [21]
11-4=7 your welcome
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3 years ago
Find the area of the figure.
Naddika [18.5K]

Answer:

55 sq meters

Step-by-step explanation:

Area of big triangle 7x5=35/2

but there's two big triangle so 35/2 x 2 =35

area of small triangle 5x4=20/2 10

10 x 2 =20. And then combine 20+35=55

So the area is 55 sq meters.

4 0
2 years ago
When an automobile is stopped with a roving safety patrol,each tire is checked for tire wear, and each headlight is checkedto se
ryzh [129]

Answer:

a) Joint ptobability distribution

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) P(X<= 1 and Y <= 1) = P(X<= 1) * P(Y<=1) = 0.56

c) P(X + Y = 0)=0.3

d) P(X + Y <= 1)=0.53

Step-by-step explanation:

We have to construct the joint probability table with the marginal probabilities of X and Y.

X can take values from 0 to 2, and Y can take values from 0 to 4.

We can calculate each point of the joint probability as:

P(x,y)=P_x(x)*P_y(y)

Then, the joint probabilities are:

X=0 Y=0 Px=0.5 Px=0.6 P(0,0)=0.3

X=0 Y=1 Px=0.5 Px=0.1 P(0,1)=0.05

X=0 Y=2 Px=0.5 Px=0.05 P(0,2)=0.025

X=0 Y=3 Px=0.5 Px=0.05 P(0,3)=0.025

X=0 Y=4 Px=0.5 Px=0.2 P(0,4)=0.1

X=1 Y=0 Px=0.3 Px=0.6 P(1,0)=0.18

X=1 Y=1 Px=0.3 Px=0.1 P(1,1)=0.03

X=1 Y=2 Px=0.3 Px=0.05 P(1,2)=0.015

X=1 Y=3 Px=0.3 Px=0.05 P(1,3)=0.015

X=1 Y=4 Px=0.3 Px=0.2 P(1,4)=0.06

X=2 Y=0 Px=0.2 Px=0.6 P(2,0)=0.12

X=2 Y=1 Px=0.2 Px=0.1 P(2,1)=0.02

X=2 Y=2 Px=0.2 Px=0.05 P(2,2)=0.01

X=2 Y=3 Px=0.2 Px=0.05 P(2,3)=0.01

X=2 Y=4 Px=0.2 Px=0.2 P(2,4)=0.04

We can write it in the form of a matrix:

\begin{pmatrix}  &Y=0&Y=1&Y=2&Y=3&Y=4\\X=0&0.3&0.05&0.025&0.025&0.1\\ X=1&0.18&0.03&0.015&0.015&0.06 \\ X=2&0.12&0.02&0.01&0.01&0.04\end{pmatrix}

b) From the joint probability P(X<= 1 and Y <= 1) is equal to

P(X\leq 1 \& Y \leq 1)=P(0,0)+P(0,1)+P(1,0)+P(1,1)\\\\P(X\leq 1 \& Y \leq 1)=0.3+0.05+0.18+0.03=0.56

We can calculate P(X<= 1) * P(Y<=1)

P_x(X\leq 1)=P_x(0)+P_x(1)=0.5+0.3=0.8\\\\P_y(Y\leq1)=P_y(0)+P_y(1)=0.6+0.1=0.7\\\\P_x(X\leq1)*P_y(Y\leq1)=0.8*0.7=0.56

Both calculations give the same result.

c) Probability of no violations

P(X+Y=0)=P(0,0)=0.3

d) P(X + Y <= 1)

P(X+Y \leq 1)=P(0,0)+P(0,1)+P(1,0)\\\\P(X+Y \leq 1)=0.3+0.05+0.18=0.53

5 0
3 years ago
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