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Anika [276]
2 years ago
15

What is the approximate distance between the points (–5, 4) and (2, –8)?

Mathematics
2 answers:
Volgvan2 years ago
8 0

Answer:

option d is the correct answer

Step-by-step explanation:

Distance Equation Solution:

d=radical(2−(−5))2+(−8−4)2

d=radical(7)2+(−12)2

d=radical49+144

d=radical 1–93

d=13.892444

Sonja [21]2 years ago
6 0
Bbfncyuvdhctbdjtcvchxu dud buddy dyvdyd buddy
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Problem page isabel runs 7 miles in 50 minutes. at the same rate, how many miles would she run in 75 minutes?
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<span>if isabel runs 7 miles in 50 minutes, then you know she runs 7/50 miles in one minute, right? So multiply that by the number of minutes she did run, and you can find the distance:


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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.

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2 years ago
An object is launched vertically in the air at 36.75 meters per second from a 6-meter-tall platform. Using the projectile motion
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I HAVE THE SAME QUESTION
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