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Olin [163]
3 years ago
15

8 cars (3 red, 3 blue, 2 yellow) are parked in a line, how many arrangements can be formed if all the cars of each colour are id

entical and the yellow cars must not be together
Mathematics
1 answer:
Ronch [10]3 years ago
4 0

Answer:

the answer is 18. There are 18 ways to arrange the cars.

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Which point on the scatter plot is an outlier?
Masja [62]

Answer:

The outlier is A) Point A

Step-by-step explanation:

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3 years ago
Geometry Homework Question ​
mash [69]

Answer:

Distance is 8.485

Midpoint is (3,4)

Step-by-step explanation:

Use the points (-1,7) and (5,1)

Insert them into the equation and solve for D

D=\sqrt{x} ((5-(-1))^2 + (1-7)^2)\\D=\sqrt{x} ((6)^2 + (-6)^2)\\D=\sqrt{x} (36+36)\\D=\sqrt{x} (72)\\D=8.485

Insert points into second equation to find midpoint

(\frac{_1+5}{2} , \frac{7+1}{2}) \\(\frac{6}{2} ,\frac{8}{2} )\\(3,4)

7 0
3 years ago
Given g(x)<br><img src="https://tex.z-dn.net/?f=g%28x%29%20%3D%200.3%20%7Bx%7D%5E%7B2%7D%20%20%2B%2052" id="TexFormula1" title="
horsena [70]

Answer:

no solution

Step-by-step explanation:

5 0
3 years ago
What is the ratio of the area of the inner square to the area of the outer square?
Anastasy [175]

Answer:

\frac{(a-b)^2+b^2}{a^2}

Step-by-step explanation:

Since, By the given diagram,

The side of the inner square = Distance between the points (0,b) and (a-b,0)

=\sqrt{(a-b-0)^2+(0-b)^2}

=\sqrt{(a-b)^2+b^2}

Thus the area of the inner square = (side)²

=(\sqrt{(a-b)^2+b^2})^2

=(a-b)^2+b^2\text{ square cm}

Now, the side of the outer square = Distance between the points (0,0) and (a,0),

=\sqrt{(a-0)^2+0^2}

=\sqrt{a^2}=a

Thus, the area of the outer square = (side)²

=a^2\text{ square cm}

Hence, the ratio of the area of the inner square to the area of the outer square

=\frac{(a-b)^2+b^2}{a^2}

4 0
3 years ago
Read 2 more answers
Eric's class consists of 12 males and 16 females. If 3 students are selected at random, find the probability that they
Reptile [31]

Answer:

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

Step-by-step explanation:

Let 'M' be the event of selecting males n(M) = 12

Number of ways of choosing 3 students From all males and females

n(M) = 28C_{3} = \frac{28!}{(28-3)!3!} =\frac{28 X 27 X 26}{3 X 2 X 1 } = 3,276

Number of ways of choosing 3 students From all males

n(M) = 12C_{3} = \frac{12!}{(12-3)!3!} =\frac{12 X 11 X 10}{3 X 2 X 1 } =220

The probability that all are male of choosing '3' students

P(E) = \frac{n(M)}{n(S)} = \frac{12 C_{3} }{28 C_{3} }

P(E) =  \frac{12 C_{3} }{28 C_{3} } = \frac{220}{3276}

P(E) = 0.067 = 6.71%

<u><em>Final answer</em></u>:-

The probability that all are male of choosing '3' students

P(E) = 0.067 = 6.71%

3 0
3 years ago
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