Ask question

Ask question

All categories

sergeinik [125]

3 years ago

5

Fifteen girls left a mixed group of boys and girls. There remained two boys for each girl. After this, 45 boys leave the group,

and there were 5 girls for each boy. Determine the number of girls in the
original group.

1 answer:

statuscvo [17]3 years ago

5 0

Sorry i cant answer this way too hard for me

You might be interested in

2. Which of these are solutions to the equation 2x3 = 54?

cricket20 [7]

The answer is C. x=3 2(3)^3=54

5 0

3 years ago

Read 2 more answers

What is the equation of the vertical line that contains the point (-3,8)?

forsale [732]

Answer: The answer is; x=-3

Step-by-step explanation: (The work is in the image below)

Additional note: hey! I hope that this helps you! Have a good day!

8 0

3 years ago

Points A, B and C lie on a circle, centre O.<br> Angle AOC = 142°.<br> Find the value of y.

aksik [14]

The value of the angle $y$ within the quadrilateral $OABC$ inscribed the circle is 142°.

Let suppose that quadrilateral $OABC$ is a parallelogram. Then, the following conditions must be fulfilled:

$\overline {OA} \cong \overline {BC}$
$\overline{AB} \cong \overline{OC}$
$\angle AOC \cong \angle ABC$

In addition, if we have a rhombus we have the additional condition as A, B and C lie on the circle:

$\overline{OA} \cong \overline {AB} \cong \overline {BC} \cong \overline{OC}$ (4)

If we know that $\angle AOC = 142^{\circ}$ , then the value of $\angle ABC = y = 142^{\circ}$ .

Based on all these finding, we conclude that the value of the angle $y$ within the quadrilateral $OABC$ inscribed the circle is 142°.

To learn more on parallelograms, we kindly invite to check this verified question: brainly.com/question/555469

5 0

2 years ago

How many solutions does this equation have?

user100 [1]

One solution

Hgyugyhvd

6 0

3 years ago

Read 2 more answers

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, <em>p</em> = 0.40.

(1)

Let <em>X</em> = 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 <em>k</em> 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 <em>Y</em> = number of computers infected.

The number of computers in the company is, <em>n</em> = 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