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2 years ago

Find the following integrals

Mathematics

2 answers:

Hunter-Best [27]2 years ago

8 0

$\large{\underline{\underline{\pmb{\sf{\color{yellow}{Answer:}}}}}}$

$\sf x - \frac{4}{x} + C$

Step-by-step explanation:

$\sf \longmapsto \int \frac{ {x}^{2} + 4 }{ {x}^{2} }dx \\ \\ \\ \sf \longmapsto \int \frac{ {x}^{2} }{ {x}^{2} } dx + \int \frac{4}{ {x}^{2} } dx \\ \\ \\ \sf \longmapsto \int 1dx + 4 \int \frac{1}{ {x}^{2} } dx \\ \\ \\ \sf \longmapsto x + 4( \frac{{x}^{ -2 + 1}}{ -2 + 1} ) + C \\ \\ \\ \boxed{ \longmapsto x - \frac{4}{x} + C}$

irina1246 [14]2 years ago

7 0

Answer:

x - 4/x

Step-by-step explanation:

Re write as 1 + 4 x^-2

then you can see the integral is x -4x^-1

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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

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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

= 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