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2 years ago

5 out of every 10 kid's meals contain a toy car. Find the probability that

Mathematics

1 answer:

tiny-mole [99]2 years ago

5 0

Answer:

Step-by-step explanation:

will you help me with my question now the one about english

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If 1 16 oz can cost. $1 how much is 8 40oz cans

3241004551 [841]

1 16oz = $1
40=16*4
1 40oz = $4

8*$4 = $32

3 0

3 years ago

Is it an identity? and how?

frutty [35]

Yes, it's an identity because remember sin 6x would be (3x + 3x)= which would make it 2 sin 3x....concluding the equation should be an identity. :)

Hoped this helped :)

6 0

3 years ago

A scientist has 100 milligrams of a radioactive element. The amount of radioactive element remaining after t days can be determi

grandymaker [24]

The mass of the first shipment at time t is
$m_{1}=100( \frac{1}{2} )^{ \frac{t}{10}}$

The mass of the second shipment at time t is
$m_{2}=100( \frac{1}{2} )^{ \frac{t-3}{10} }$

At time t, the ratio of m₁ to m₂ is
$\frac{m_{1}}{m_{2}} = \frac{100}{100}. \frac{(1/2)^{t/10}}{(1/2)^{(t-3)/10}} \\ = \frac{(1/2)^{t/10}}{(1/2)^{t/10}}. \frac{1}{(1/2)^{-3/10}} \\ = (1/2)^{3/10} \\ = 0.8123$

Therefore as a percentage,
$\frac{m_{1}}{m_{2}} =100*0.8123 = 81.23 \%$

Answer: B. 81.2%

4 0

4 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

4 years ago

Determine the percentage of the square that belongs to each of the four regions. (image included)

tatyana61 [14]

Step-by-step explanation:

maybe you can try to measure each of the ways maybe and you can get the equation

3 0

2 years ago