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wlad13 [49]
2 years ago
15

A small class has 10 students, 5 of whom are girls and 5 of whom are boys. The teacher is going to choose two of the students at

random. What is the
probability that the first student chosen will be a girl and the second will be a boy? Write your answer as a fraction in simplest form?
Mathematics
2 answers:
IgorLugansk [536]2 years ago
7 0

Answer: 1/5

Step-by-step explanation:

user100 [1]2 years ago
4 0

Answer:

the chances are most likely going to be 50/50

Step-by-step explanation:

<u>50</u>

50

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horsena [70]
The answer is C) 0.03, 0.065, 1/8, 14%
5 0
2 years ago
if a real estate company gives 3% commission on to it's agents. if an agent sold a piece of land for 12,50,000, how much cccommi
liubo4ka [24]

Answer:

37500

Step-by-step explanation:

3% of 1250000

That is (3/100)*1250000

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4 0
3 years ago
The indefinite integral can be found in more than one way. First use the substitution method to find the indefinite integral. Th
Fantom [35]

Answer:

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C

Step-by-step explanation:

To find:

∫6x^5(x^6-2)\,dx

Solution:

Method of substitution:

Let x^6-2=t

Differentiate both sides with respect to t

6x^5\,dx=dt

[use (x^n)'=nx^{n-1}]

So,

∫6x^5(x^6-2)\,dx = ∫ t\,dt = \frac{t^2}{2}+C_1 where C_1 is a variable.

(Use ∫t^n\,dt=\frac{t^{n+1} }{n+1} )

Put t=x^6-2

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C_1

Use (a-b)^2=a^2+b^2-2ab

So,

∫6x^5(x^6-2)\,dx = \frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C

where C=2+C_1

Without using substitution:

∫6x^5(x^6-2)\,dx = ∫6x^{11}-12x^5\,dx = \frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C

So, same answer is obtained in both the cases.

7 0
2 years ago
Can anyone help with expanding this?
igomit [66]

Answer:

log(4) +3 log(x) +7 log(y)

Step-by-step explanation:

Hope it helps!

7 0
3 years ago
Read 2 more answers
Is it possible for a logarithm to equal zero
tatyana61 [14]
Yes it is. log(1) and ln(1) both equal zero. But you cannot take the logarithm of 0.
6 0
3 years ago
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