9 < 11, so 9(5) < 11(5) true or false?

List at least five importance of having self discipline m following protocol during enhanced community quarantine in your respec

Maksim231197 [3]

Answer:

no clue -0- if I knew I would tell you

4 0

3 years ago

On April 1, Preston McCord deposited $1,400 in a savings account that pays annual interest of 3.2%

lesya692 [45]

Answer:

After 6 months, the interest generated by the investment would be $21.10.

Step-by-step explanation:

To determine the interest that Preston McCord could earn by investing $ 1,400 in an account that pays 3.2% annual interest compounded monthly, leaving said money invested for a period of 6 months, it is necessary to perform the following calculation:

X = 1,400 x (1 + 0.032 / 6) ^ 0.5x6

X = 1,421.10

1421.10 - 1400 = 21.10

Thus, after 6 months, the interest generated by the account will be $ 21.10.

4 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 <em>k</em> objects from <em>n</em> 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

Simplify the ratio: 2:1 1/2

yKpoI14uk [10]

1/2 is the simplified version

8 0

3 years ago

(15 points)

boyakko [2]

B. is the Correct Answer, Hope It Helps!

3 0

4 years ago

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