21 fa cumulative distribution function for a continuous random variable is deflhed By. %3D x < 1 x3 + x2 F(x) = 2 0< x < 1 Find the density for X

Q: f c = 12, what is 4 + c?

A: given,  c=12 let  A=4+c

Q: 5. If a number plus 4 equals two times 6, what is the number?

A:

Q: so what is the solution

A: Given N=120 P=66% X=81 We have to calculate probability that at most 81

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A:

Q: Can you clarify the answer for the second step (a) ? It says that it's wrong.

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Q: That is NOT the correct solution. Thanks.

A:

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A:

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A: given equations are…

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A: Hello! Thank you for the question. We'll able to solve only one question as per our honor code.

Q: х - 12 1 18

A: Solving the problem by finding the value of x.

Q: 18 × 8 y – 4 = 24

A: The objective is to find the value of y

Q: You gave me the same answer for two different questions with different values?

A:

Q: general solution

A:

Q: 2х + 1 5x -11 4у + 5 9у — 10

A:

Q: these ans

A: NoItems Time 1 4.8 1 14.7 1 13.7 1 9.5 1 2.4 1 11.8 2 10.5 2 12 2 14.3 2 16.4…

Q: Solve h=ab+rh=ab+r for a

A: Given     h= ab + rh We can solve for a by separating it

Q: For step three, how did you replace the h?

A: Let's solve.

Q: Solve for r. 13

A:

Q: could you actually show the work for the answer?

A: Given that,the equation of the tangent line to fx at the point a=1 is y=7-2x.We need to find the…

Q: solution is typed off the page

A:

Q: Explain again step 3 ? I don't understand.

A: It is clear that the function f(t) has jump discontinuity at (n,0) where n is an integer.

Q: Where is the answer and solution for H0 and H1?

A: The mean SAT score in mathematics is 486.

Q: How do you get those answers?

A: 3-point throw  n= 3 Possible outcomes : 3 free throw ,  0 missing = (3,0) 2 free throw , 1 missing =…

Q: 12 C 3 В 9

A:

Q: (15x - 8) 226

A: Given,

Q: ADEFAWVU 18x+6 D 156

A:

Q: | (i) On the interval [3, 10], f: does attain an absolute maximum; does not attain an absolute…

A: In the interval [3,10] , on the right side of x=5 the graph goes to its highest position. But there…