Find the missing side link when the area is 171 km².

Twice the sum of a number and 3 is equal to three times the difference of the number and 6. Find the number.

vfiekz [6]

Answer: 5

Step-by-step explanation:

7 0

2 years ago

Please answer it now in two minutes

o-na [289]

Answer:

p = 3

Step-by-step explanation:

This is an isosceles right triangle

=> 2side^2 = (hypotenuse)^2

=> 2p^2 = (3sqrt(2))^2

=> 2p^2 = 18

=> p^2 = 9

=> p = 3

5 0

3 years ago

The center of a circle is located at (3, 8), and the circle has a radius that is 5 units long. What is the general form of the e

lana [24]

First we write equation that consist coordinates of center and radius. That formula goes like this:
(x-x1)^2 + (y-y1)^2 = r^2

x and y are coordinates on any point on circle
x1 and y1 are coordinates of center of circle.
r is radius of that circle. now we need to express our values for center and radius andf square binoms and see what matches in our options.

(x-3)^2 + (y-8)^2 = 5^2
x^2 + y^2 -6x -16y +48 = 0

The answer is first option.

5 0

3 years ago

In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is

Alex777 [14]

Answer:

(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2) The probability at least 5 computers are infected is 0.949.

Step-by-step explanation:

The probability that a computer is defective is, p = 0.40.

(1)

Let X = number of computers to be tested before the 1st defect is found.

Then the random variable $X\sim Geo(p)$ .

The probability function of a Geometric distribution for k failures before the 1st success is:

$P (X = k)=(1-p)^{k}p;\ k=0, 1, 2, 3,...$

Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:

P (X ≥ 5) = 1 - P (X < 5)

= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

$=1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078$

Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2)

Let Y = number of computers infected.

The number of computers in the company is, n = 20.

Then the random variable $Y\sim Bin(20,0.40)$ .

The probability function of a binomial distribution is:

$P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...$

Compute the probability at least 5 computers are infected as follows:

P (Y ≥ 5) = 1 - P (Y < 5)

= 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)] $=1-[{20\choose 0}(0.40)^{0}(1-0.40)^{20-0}+{20\choose 1}(0.40)^{1}(1-0.40)^{20-1}\\+{20\choose 2}(0.40)^{2}(1-0.40)^{20-2}+{20\choose 3}(0.40)^{3}(1-0.40)^{20-3}\\+{20\choose 4}(0.40)^{4}(1-0.40)^{20-4}]\\=1-[0.00004+0.00049+0.00309+0.01235+0.03499]\\=1-0.05096\\=0.94904$

Thus, the probability at least 5 computers are infected is 0.949.

7 0

3 years ago

Ashley and a friend have dinner at Red Lobster. Ashley’s meal costs $15.85 and she leaves an 18% tip. Her friend’s meal costs $1

FinnZ [79.3K]

Answer: Ashley spent $18.70 and the friend spent $17.16

Step-by-step explanation:

Original cost of Ashley's meal = $ 15.85

tip left = 18% of 15.85

that is

tip left = 0.18 x 15.85 = $2.85

All together, Ashley will pay

$15.85 + $2.85

= $18.70

The friend's original cost = $14.30

tip = 20% of 14.30

= 0.20 x 14.30

= $2.86

All together , the friend will pay

$14.30 + $2.86

=$17.16

Therefore:

Ashley spent $18.70 and the friend spent $17.16

4 0

2 years ago