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jenyasd209 [6]
3 years ago
10

Helpppppppppppppppppppppppppp

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
7 0
Here is the answer
Hope it helps

IgorLugansk [536]3 years ago
6 0

Answer:

Step-by-step explanation:

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Order the following from least to greatest.<br> 3.1, 118%, 5/12, 0.06, 2/7
photoshop1234 [79]

Answer:

0.06, 2/7, 5/12, 118%, 3.1

Step-by-step explanation:

7 0
3 years ago
Please help tanks i could really use i need to graduate on time
Hitman42 [59]

Answer:

A

Step-by-step explanation:

7 0
4 years ago
A denotes an mn matrix. Determine whether the statement is true or false. Justify your answer. The row space of AT is the same a
ziro4ka [17]

Answer: true

Step-by-step explanation:

For an m*n matrix, the column space of A will be a space formed by the lineal combination of all the columns of A.

column space = a1*c1 + a2*c2 + ...

                       

where a1, a2, ... are scalars, and c1 is the vector of column 1.

Then we should write:

Column space = a1*(A₁₁, A₂₁, A₃₁, ...) + a2*(A₁₂, A₂₂, A₃₂, ...) + ...

Now, the transpose is defined as:

[At]₁₃ = A₃₁

Here i used the element with subindex 3 and 1, but is the same for every subindex.

Notice that if A is m*n, then [At] is n*m

Now, the row space of [At] will be, same as before.

Row space = b1*r1 + b2*r2 + ...

Where b1, b2, ... are scalars and the r's are the vector of each row.

                   = b1*( [At]₁₁ , [At]₁₂, [At]₁₃, ...) + b2*([At]₂₁, [At]₂₂, [At]₂₃, ...) + ...

Now we replace each term of the transpose by the associated element in the original matrix.

                   = b1*( A₁₁, A₂₁, A₃₁, ...) + b2*(A₂₁, A₂₂, ...) + ....

If we take:

b1 = a1, b2 = a2, b3 = a3, ...

We will have that the row space of [At] is the same as the column space of A.

4 0
3 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
Given: 5x + 3 &gt; 4x + 7.<br><br><br><br> Choose the graph of the solution set.
Natali5045456 [20]

Answer:

5x+3>4x+7

-4x      -4x

x+3>7

-3     -3

x>4

so the answer will be the second

Step-by-step explanation:

7 0
3 years ago
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