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Pavlova-9 [17]
3 years ago
11

The digits 4 comma 5 comma 6 comma 7 comma and 8 are randomly arranged to form a​ three-digit number. ​ (Digits are not​ repeate

d.) Find the probability that the number is even and greater than 800.
Mathematics
1 answer:
fenix001 [56]3 years ago
3 0

Answer:

0.1 or 1/10

Step-by-step explanation:

The digits 4, 5, 6, 7 and 8 are randomly arranged to form a three digit number, where the digits are not repeated.

This is question of permutation.

Imagine this sum as; there are 3 boxes(blank spaces for digits) and 5 different fruits(digits) are to be put in these boxes, where a box can hold a maximum of only 1 fruit. The number of such permutations are: ⁵P₃

By formula (a! is factorial a):

ᵃPₙ = \frac{a!}{(a-n)!}

⁵P₃ = \frac{5!}{(5-3)!}

⁵P₃ = \frac{5!}{2!}

⁵P₃ = \frac{5*4*3*2*1}{2*1}

⁵P₃= 60

This is the total count of possible numbers that can be formed.

Now, for a number to be greater than 800 and even; first digit should necessarily be 8. Last digit can be 4 or 6. Using these conditions, there are 6 possibilities. 854, 864, 874, 846, 856, 876 are the numbers.

The probability that number is even and greater than 800 is:

P=\frac{ favorable outcomes}{equally likely possible outcomes}

P=\frac{6}{60}

P=\frac{1}{10}

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Complete Question:

The complete question is shown on the first uploaded image

Answer:

a) The graph of  f(y) versus y. is shown on the second uploaded image

b) The critical point is at y = 0  and the solution is asymptotically unstable.

c)The phase line is shown on the third uploaded image

d) The sketch for the several graphs of solution in the ty-plane  is shown on the fourth uploaded image

Step-by-step explanation:

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Looking at the given differential equation

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