How to you write a rule for 2, 10, 5 and 250?

Whats the answer to this problem???? the product of 6 and a number p is 24

Ad libitum [116K]

Well the P should probably be 4

7 0

3 years ago

3x^4-18x^3+9x^2-54x Ax (x^2+B) (x+C) What is the value of C

lesantik [10]

Well to do this correctly it would have to be a but not techonology.

4 0

3 years ago

What is 4.284 rounded to 2 significant digits.

kolezko [41]

Step-by-step explanation:

4.284

≈4.3

Hope it helps ya

3 0

3 years ago

Mr. Gold had 4 times more cars than motorcycles in his garage. After he bought 4 more cars and sold 4 motorcycles, there were 8

Semenov [28]

Answer: There are 40 cars and 5 motorcycles at the end of buying and selling. He started with 9 motorcycles and 36 cars.

Step-by-step explanation:

x=# of motorcycles 4x=# of cars

8(x-4)=4x+4

8x-32=4x+4

4x=36

x=9

---

4(9)+4=36+4=40

9-4=5

8 0

3 years ago

Read 2 more answers

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:

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

Thus, there are 70 different ways set up 4 computers out of 8.

(b)

It is provided that 4 computers are randomly selected.

Compute the probability that exactly 3 of the 4 computers selected are desktops as follows:

$P(Y=3)={4\choose 3}\times(\frac{5}{9})^{3}\times (1-\frac{5}{9})^{4-3}\\=4\times\frac{125}{729}\times\frac{4}{9}\\ =0.304832\\\approx0.305$

Thus, the probability that exactly three of the selected computers are desktops is 0.305.

(c)

Compute the probability that of the 4 computers selected at least 3 are desktops as follows:

$P(Y\geq 3)=1-P(Y$

Thus, the probability that at least three of the selected computers are desktops is 0.401.

6 0

3 years ago