HELP The perimeter of a rectangle is 150 inches the lengh is 3 inches longer than 5times the width find the length

klasskru [66]

I would make a tree because there are not that many possibilities. to get 99 with numbers between 1 and 50 you have 50+49 and 49 and 50 nothing else will work because if one spin is less than 49 say 48 to get 99 you need 51 which is impossible. either you 50 then 49 or you get 49 then 50. the others are a bit harder because you need to find all possible spins that will sum to 52 which is a much larger number.

8 0

3 years ago

Triangle a' B' c' is dilation of triangle abc about point p. Is this a reduction or enlargement

Yuliya22 [10]

Enlargement

Hope this helps :)

8 0

3 years ago

The computers of nine engineers at a certain company are to be replaced. Four of the engineers have selected laptops and the oth

Gala2k [10]

Answer:

(a) There are 70 different ways set up 4 computers out of 8.

(b) The probability that exactly three of the selected computers are desktops is 0.305.

(c) The probability that at least three of the selected computers are desktops is 0.401.

Step-by-step explanation:

Of the 9 new computers 4 are laptops and 5 are desktop.

Let X = a laptop is selected and Y = a desktop is selected.

The probability of selecting a laptop is = $P(Laptop) = p_{X} = \frac{4}{9}$

The probability of selecting a desktop is = $P(Desktop) = p_{Y} = \frac{5}{9}$

Then both X and Y follows Binomial distribution.

$X\sim Bin(9, \frac{4}{9})\\ Y\sim Bin(9, \frac{5}{9})$

The probability function of a binomial distribution is:

$P(U=k)={n\choose k}\times(p)^{k}\times (1-p)^{n-k}$

(a)

Combination is used to determine the number of ways to select k objects from n distinct objects without replacement.

It is denotes as: ${n\choose k}=\frac{n!}{k!(n-k)!}$

In this case 4 computers are to selected of 8 to be set up. Since there cannot be replacement, i.e. we cannot set up one computer twice or thrice, use combinations to determine the number of ways to set up 4 computers of 8.

The number of ways to set up 4 computers of 8 is:

${8\choose 4}=\frac{8!}{4!(8-4)!}\\=\frac{8!}{4!\times 4!} \\=70$