Help, i need an answer for my test

Mara found the length of time of an investment. The principal of the investment was $4,300, the interest rate was 6.2 percent, a

77julia77 [94]

well, 6.2% of 4,300 can't be over half. she has made a calculation error, and the real answer is (this is where the mistake was made) 4,300 * 0.062 (not 6.2, because percents are always divided by 100 to make them into a usable number), which is equal to $266.60.

3 0

3 years ago

a) Everyone on the team talks until the entire team agrees on one decision. O b) Everyone on the team discusses options and then

insens350 [35]

Answer:

a) Everyone on the team talks until the entire team agrees on one decision.

Step-by-step explanation:

Option B consists of voting and not everyone would like the outcome. Option C is making an outsider the decision maker, which can't be helpful since he / she won't have as strong opinions as the team itself. Option D is just plain wrong as it defeats the purpose of team work and deciding as one team. So, I believe option A makes the most sense

8 0

3 years ago

Help me with hanger math...

andrew11 [14]

The answer is 3 First you subtract 2 from 17 and then you divide 15 by 5 to get the answer of 3

6 0

3 years ago

PLZ HELP ME ITS FOR ALGEBRA

oksano4ka [1.4K]

Answer:

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Step-by-step explanation:

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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 <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