Given:
Cards labelled 1, 3, 5, 6, 8 and 9.
A card is drawn and not replaced. Then a second card is drawn at random.
To find:
The probability of drawing 2 even numbers.
Solution:
We have,
Even number cards = 6, 8
Odd numbers cards = 1, 3, 5, 9
Total cards = 1, 3, 5, 6, 8 and 9
Number of even cards = 2
Number of total cards = 6
So, the probability of getting an even card in first draw is:



Now,
Number of remaining even cards = 1
Number of remaining cards = 5
So, the probability of getting an even card in second draw is:


The probability of drawing 2 even numbers is:



Therefore, the probability of drawing 2 even numbers is
. Hence, the correct option is (b).
Answer:
Step-by-step explanation:
Given that the owner of a motel has 2900 m of fencing and wants to enclose a rectangular plot of land that borders a straight highway.
Fencing is used for 2times length and 1 width if highway side is taken as width
So we have 2l+w = 2900
Or w = 2900-2l
Area of the rectangular region = lw

Use derivative test to find the maximum

So maximum when I derivative =0
i.e when 
Largest area = A(725)
= 
1051250 sqm is area maximum
Yes, it is true.
________________
For example,
take two integers 12 and 4
12 × 4 = 48 : product is positive
12 ÷ 4 = 3: quotient is positive
____________________
If you take -12 and 4
-12 × 4 = -48 : product is negative
-12 ÷ 4 = -3: quotient is negative
Answer:
Jermey is not correct because abouslte value is always postive
Step-by-step explanation:
Hello from MrBillDoesMath!
Answer:
Solutions: x = +\- 5i or x = +\- sqrt(5)
Discussion:
Factor x^4 - 25:
x^4 - 25 = (x^2+5) (x^2-5) => factor x^2 - 5
x^4 - 25 = (x^2+5)(x + sqrt(5)) (x - sqrt(5)) => factor x^2 + 5
x^4 = 25 = (x +5i)(x-5i) (x + sqrt(5)) (x - sqrt(5))
Hence the solutions are
x = +\- 5i and x = +\- sqrt(5)
Thank you,
MrB