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brilliants [131]
3 years ago
7

PLEASE HELP QUICK !!!!!

Mathematics
1 answer:
Semmy [17]3 years ago
8 0

Answer:

the value of P(B) = 1/3

Step By Step Explaination:

If A and B are two independent events, then P(A and B)=P(A) x P(B)  (i)

Put all values in (i), we get

1/6= 1/2  x P(B)

P (B) = 2/6

P(B) = 1/3

Hence The Value of P(B) = 1/3.

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The equation of the line is
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How to expand and simplify this equation 7(2a+3)+3(4a-2).
Pavel [41]
Ok, so first we distribute, you multiply the seven into everything in the parentheses next to it. So far we have, 14a+21+3(4a-2). You distribute the three into the parentheses to get, 14a+21+12a-6. You combine the like terms to get, 26a-15. You cannot simplify it any further so the answer is 26a-15.
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2 years ago
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Which ratio is porportional to 18/21? A.2/3 B. 30/35 C. 36/48 D.1/3
defon
You can solve this problem by applying the proccedure below:
 
 You have:
 
 a) 18/21=2/3
 18x3/21x2
 36≠42 (2/3 is not proportinal to 18/21).
 
 b)18/21=30/35
 18x35/21x30
 630=630  (30/35 is proportional to 18/21)
 
 c)18/21=36/48
 18x48/21x36
 864≠756  (36/48<span> is</span><span> not proportinal to 18/21)
</span> 
 d)18/21=1/3
 18x3/21x1
 54≠21  (<span> 1/3 is</span><span> not proportinal to 18/21)
</span> 
 Therefore, you can conclude that the correct option is: B.30/35
8 0
2 years ago
Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators
Nata [24]

Answer:

The solution is y=-\frac{12}{12x-30x^2+1}.

Step-by-step explanation:

A first order differential equation y'=f(x,y) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y:

f(x,y)=p(x)h(y)

where p(x) and h(y) are continuous functions.

We have the following differential equation

y'=(1-5x)y^2, \quad y(0)=-12

In the given case p(x)=1-5x and h(y)=y^2.

We divide the equation by h(y) and move dx to the right side:

\frac{1}{y^2}dy\:=(1-5x)dx

Next, integrate both sides:

\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C

Now, we solve for y

-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}

We use the initial condition y(0)=-12 to find the value of C.

-12=-\frac{2}{2\left(0\right)-5\left(0\right)^2+2C}\\-12=-\frac{1}{c}\\c=\frac{1}{12}

Therefore,

y=-\frac{2}{2x-5x^2+2(\frac{1}{12})}\\y=-\frac{12}{12x-30x^2+1}

4 0
3 years ago
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