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nignag [31]
2 years ago
8

Which system of equations has no solutions? y - ​

Mathematics
1 answer:
Alona [7]2 years ago
7 0
The second one on the top right
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Please factorise 16p²-9q² for me<br>​
kirill115 [55]

Answer:

a ^ 2 - b ^ 2 = ( a + b ) ( a - b ) where a = 4p and b = 3q.

( 4p + 3q ) ( 4p - 3q )

7 0
2 years ago
A large but sparsely populated county has two small hospitals, one at the south end of the county and one at the north end. The
Readme [11.4K]

Answer:

Step-by-step explanation:

Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by

f(x,y) = P((X=x,Y=y) ,

and given that the random variables are independent the joint pmf isbgiven by:

f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)

(b) the required probability is given by considering X=0,1 and Y =0,1

f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }

= [0.1 +0.2 ] × [0.1 + 0.3]

= 0.3 × 0.4

= 0.12

Now we find

fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1

fX(x) = sum{[f(x,y)]}= 0.1+0.3

= 0.4

fY(y) = sum{[f(x.y)]} = 0.1 + 0.2

= 0.3

Since fY(y) × fX(x) = 0.4 × 0.3

= 0.12

Hence f(x,y) = fX(x) .fY(y), the events are independent

(c) the required event is give by: [P(X<=1, PY<=1)]

= P(X<=1) . P(Y<=1)

= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1

= 0.4 × 0.3

= 0.12

(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A

A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]

A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]

A = [0.09] + [ 0.09]

A = 0.18

Pls note the sum of the vertical column X=2 is 0.3

3 0
3 years ago
What are the steps to solve this problem?
adoni [48]
I'm sorry you need to be more clear I don't understand the question.
7 0
3 years ago
Just #2 . Help !! Please :)
kirza4 [7]
180- 113 = 67

180- 67= 113

<5= 67
<2=113
8 0
3 years ago
Read 2 more answers
Ln(1/e^2) without calculator
castortr0y [4]
Once you remember the definition of a log, the answer to this question will literally fall out of your pencil.

First, ' Ln ' means 'natural log' ... logs to the base of ' e '.

Definition of the natural log of a number:
In order to get the number, what power do I have to raise ' e ' to ?

OK.  What power do you have to raise ' e ' to in order to get  1/e²  ?

Isn't  1/e²  the same thing as  e⁻²  ?

So, in order to get  1/e² , you have to raise ' e ' to the  -2  power .

In math-speak:    Ln(1/e²) = <em><u>-2</u></em> .
7 0
2 years ago
Read 2 more answers
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