A card is chosen from a random deck, then a month of the year is chosen. Find the probability of getting a face card and June.

Mathematics

1 answer:

gogolik [260]2 years ago

3 0

Answer: Probability of chossing a face card and the month of june = 0.019

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Find the indicated probability. round to three decimal places. a test consists of 10

Charra [1.4K]

To answer this problem, we use the binomial distribution formula for probability:

P (x) = [n! / (n-x)! x!] p^x q^(n-x)

Where,

n = the total number of test questions = 10

x = the total number of test questions to pass = >6

p = probability of success = 0.5

q = probability of failure = 0.5

Given the formula, let us calculate for the probabilities that the student will get at least 6 correct questions by guessing.

P (6) = [10! / (4)! 6!] (0.5)^6 0.5^(4) = 0.205078

P (7) = [10! / (3)! 7!] (0.5)^7 0.5^(3) = 0.117188

P (8) = [10! / (2)! 8!] (0.5)^8 0.5^(2) = 0.043945

P (9) = [10! / (1)! 9!] (0.5)^9 0.5^(1) = 0.009766

P (10) = [10! / (0)! 10!] (0.5)^10 0.5^(0) = 0.000977

Total Probability = 0.376953 = 0.38 = 38%

There is a 38% chance the student will pass.

7 0

2 years ago

5x +1 = -3+10x how do you solve this

Natali5045456 [20]

Answer:

x = 4/5

Step-by-step explanation:

Group like terms together. Start by subtracting 1 from both sides, obtaining

5x = -4 + 10x.

Then combine the x terms, obtaining 4 = 5x.

Solving for x by dividing both sides by 5, we get x = 4/5.

7 0

3 years ago

Read 2 more answers

Solve the proportion

mario62 [17]

Answer:

$\boxed{\sf x = 5}$

Step-by-step explanation:

$\sf Solve \: for \: x \: over \: the \: real \: numbers: \\ \sf \implies \frac{2}{x - 3} = \frac{5}{x} \\ \\ \sf Take \: the \: reciprocal \: of \: both \: sides: \\ \sf \implies \frac{x - 3}{2} = \frac{x}{5} \\ \\ \sf Expand \: out \: terms \: of \: the \: left \: hand \: side: \\ \\ \sf \implies \frac{x}{2} - \frac{3}{2} = \frac{x}{5} \\ \\ \sf Subtract \: \frac{x}{5} - \frac{3}{2} \: from \: both \: sides: \\ \sf \implies \frac{x}{2} - \frac{3}{2} - ( \frac{x}{5} - \frac{3}{2} ) = \frac{x}{5} - ( \frac{x}{5} - \frac{3}{2} ) \\ \\ \sf \implies \frac{x}{2} - \frac{3}{2} - \frac{x}{5} + \frac{3}{2} = \frac{x}{5} - \frac{x}{5} + \frac{3}{2} \\ \\ \sf \frac{x}{5} - \frac{x}{5} = 0 : \\ \sf \implies \frac{x}{2} - \frac{x}{5} - \frac{3}{2} + \frac{3}{2} = \frac{3}{2} \\ \\ \sf \frac{3}{2} - \frac{3}{2} = 0: \\ \sf \implies \frac{x}{2} - \frac{x}{5} = \frac{3}{2} \\ \\ \sf \frac{x}{2} - \frac{x}{5} = \frac{5x - 2x}{10} = \frac{3x}{10} : \\ \sf \implies \frac{3x}{10} = \frac{3}{2} \\ \\ \sf Multiply \: both \: sides \: by \: \frac{10}{3} : \\ \sf \implies \frac{3x}{10} \times \frac{10}{3} = \frac{3}{2 } \times \frac{10}{3} \\ \\ \sf \frac{3x}{10} \times \frac{10}{3} = \cancel{\frac{3}{10} } \times( x) \times \cancel{ \frac{10}{3} } = x : \\ \sf \implies x = \frac{3}{2} \times \frac{10}{3} \\ \\ \sf \frac{3}{2} \times \frac{10}{3} = \cancel{ \frac{3}{2} } \times \cancel{ \frac{3}{2} } \times 5 : \\ \sf \implies x = 5$