The probability that the tenth person randomly interviewed in that city is the fifth one to own a dog = 0.051
Given,
Consider the interview of a person is a trial.
Lets consider it success if a person owns a dog and consequently , a person without a dog will be a failure.'
Let the random variable X represent the number of persons to be interviewed to get 5 dog owner: in other words, to get 5 successors.
Probability of a success in each trial is p = 0.3. Therefore probability of a failure in each trial is q = 1 - p = 0.7 .
Because trials are independent. X has a negative binomial distribution with parameters k = 5, p = 0.3
P(X =x ) = b(x: k, p) = ![[(x-1)(k-1)] p^kq^r^-^k](https://tex.z-dn.net/?f=%5B%28x-1%29%28k-1%29%5D%20p%5Ekq%5Er%5E-%5Ek)
where, x = k , k + 1 , k + 2......(1)
Now, Lets find the probability that the tenth person randomly interviewed in that city is the fifth one to own a dog
Using equation (1), we get:
P(X = 10) = b(10: 5, 0.3)
= [(10-1) (5 - 1)]
= [(9)(4)]
= 0.05146
Hence, The probability that the tenth person randomly interviewed in that city is the fifth one to own a dog = 0.051
Learn more about Probability at:
brainly.com/question/743546
#SPJ9
1.8, Problem 37: A lidless cardboard box is to be made with a volume of 4 m3
. Find the
dimensions of the box that requires the least amount of cardboard.
Solution: If the dimensions of our box are x, y, and z, then we’re seeking to minimize
A(x, y, z) = xy + 2xz + 2yz subject to the constraint that xyz = 4. Our first step is to make
the first function a function of just 2 variables. From xyz = 4, we see z = 4/xy, and if we substitute
this into A(x, y, z), we obtain a new function A(x, y) = xy + 8/y + 8/x. Since we’re optimizing
something, we want to calculate the critical points, which occur when Ax = Ay = 0 or either Ax
or Ay is undefined. If Ax or Ay is undefined, then x = 0 or y = 0, which means xyz = 4 can’t
hold. So, we calculate when Ax = 0 = Ay. Ax = y − 8/x2 = 0 and Ay = x − 8/y2 = 0. From
these, we obtain x
2y = 8 = xy2
. This forces x = y = 2, which forces z = 1. Calculating second
derivatives and applying the second derivative test, we see that (x, y) = (2, 2) is a local minimum
for A(x, y). To show it’s an absolute minimum, first notice that A(x, y) is defined for all choices
of x and y that are positive (if x and y are arbitrarily large, you can still make z REALLY small
so that xyz = 4 still). Therefore, the domain is NOT a closed and bounded region (it’s neither
closed nor bounded), so you can’t apply the Extreme Value Theorem. However, you can salvage
something: observe what happens to A(x, y) as x → 0, as y → 0, as x → ∞, and y → ∞. In each
of these cases, at least one of the variables must go to ∞, meaning that A(x, y) goes to ∞. Thus,
moving away from (2, 2) forces A(x, y) to increase, and so (2, 2) is an absolute minimum for A(x, y).
Answer:
- 12k² + kl
Step-by-step explanation:
Given
7kl - 3k(4k + 2l) ← distribute parenthesis by - 3
= 7kl - 12k² - 6kl ← collect like terms
= - 12k² + kl
the solution is here,
the coordinate of centre of circle A is (3,2)
and the coordinate of centre of circle B is (3,0).
so the translation point from A to B is (3-0,0-2)=(3,-2).
Now the translation rule is given by (x+h,y+k)
where h is the tranalation anlong the x-axis and k is the translation along the y-axis.
For this problem, translation ruleis (x+3,y-2).
Then, radius of circle A(r1)=2 units
radius of circle B (r2)= 3 units
as the circle A is translated to B, the scale factor is
r2/r1=3/2=1.5
In conclusion, the translation rule for given circles is (x+3,y-2) and its scale factor is 1.5
