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3 years ago

sarah lives 7.15 kilometers away from her school. which fraction represents the distance from sarah’s house to her school

Mathematics

1 answer:

11Alexandr11 [23.1K]3 years ago

5 0

Answer:

B, 7 3/20

Step-by-step explanation:

This is a mixed fraction because it has a whole number. The 7 is put to the side. The .15 is put over 100 and makes the fraction 15/100. This can be simplified into 3/20 by dividing by 5.

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Question:

What is the m∠ACB in the circle shown below?

Answer:

Step-by-step explanation:

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2 years ago

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3 years ago

What can be multiplied to get 5 and added to get 4

bonufazy [111]

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3 years ago

PLEASE HELP, WILL GIVE BRAINLIEST!!! Find the inverse of f(x)=6x-4 and find f^-1(62)

Lena [83]

Answer: 11

=========================

Explanation:

To find the inverse, or f^-1 of a function, you first replace the y= 6x-4 with x=6y-4

Finding the inverse from here is basically just finding y in terms of x.

1) x=6y-4
2) x+4=6
3) (x+4)/6= y

y= (x+4)/6 is now your f^-1.

Now that you have your inverse, replace 62 with your x in y=(x+4)/6

1) (62+4)/6
2) 66/6
3) 11

Therefore, f^-1(62)= 11

Hope this helps :)

5 0

2 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.

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