Mathematics, 01.03.2020 06:26 allytrujillo20oy0dib

A sample of 3 different calculators is randomly selected from a group containing 17 that are bad and 36 that are good. What is the probability that all three calculators are good?

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Answer from: emmaguentherp3hjd3

Probability of all 3 calculators being good = 0.971.

Step-by-step explanation:

We have , A sample of 3 different calculators is randomly selected from a group containing 17 that are bad and 36 that are good. In order to calculate that all 3 randomly selected calculators are good , let's calculate probability of all 3 being bad and subtracting value by 1. i.e. Probability of all 3 being good = 1 - Probability of all 3 being bad

⇒ Probability of all 3 being bad = $\frac{17}{53}$ × $\frac{17}{53}$ × $\frac{17}{53}$ = 4193/148877

∴Probability of all 3 being good = 1 - Probability of all 3 being bad

⇒ $1 - \frac{4193}{148877}$

⇒ $\frac{144684}{148877}$

∴Probability of all 3 calculators being good = 0.971.

Answer from: Quest

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

Answer from: Quest

$a_1 = 1$ , and

$a_{n+1} = a_n + (n + 1)$ for $n \ge 1$

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote $a_1 = 1$ , $a_2 = 3$ , $a_3 = 6$ , so that we can call

1 the 1st term, 3 the 2nd term, 6 the 3rd term,etc.

note the following:

from the 1st term (1) to the 2nd term (3), we go up by 2,from the 2nd term (3) to the 3rd term (6), we go up by 3from the 3rd term (6) to the 4th term (10), we go up by 4, andfrom the 4th term (10) to the 5th term (15), we go up by 5.

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from $a_1 = 1$ , to get to $a_2 = 3$ , we add $2(=1+1)$

from $a_2 = 3$ , to get to $a_3 = 6$ , we add $3 (=2+1)$

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

$a_1 = 1$ , and $a_{n+1} = a_n +(n + 1)$ for $n \ge 1$

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