Which quotient is reasonable? Do not perform the division. 7315 ÷ 12 A. 6 R 9 B. 60 R 9 C. 609 D. 6,009

Mathematics

1 answer:

ANTONII [103]3 years ago

5 0

Answer:

C. 609

Step-by-step explanation:

In the question, 4-digits number were given to be divided by a 2-digits number. Without performing the division, it is only reasonable to say that the quotient will be a 3-digits number.

In the given options, the only 3-digits number is 609

Also, you can multiply 609 by 12 to check how reasonable your guess is.

⇒ 609 x 12 = 7308

The difference between 7315 and 7308 is 7.

The remaining number (7) is not up to 12 and when 7 is divided by 12 it will not give a whole number digit.

Therefore, the only reasonable quotient in the given options is 609.

You might be interested in

A recent college graduate is planning to take the first three actuarial examinations in the coming summer. She will take the fir

erica [24]

Answer:

(a) 0.7

(b) $\frac 13$

Step-by-step explanation:

Let $E_1, E_2,$ and $E_3$ be the events of passing the 1st, 2nd, and 3rd exam individually.

Given that $P(E_1)=0.9\;\cdots(i)$

where $P(E_1)$ denotes the probability of passing the 1st exam.

Condition for passing the second exam is, at first, the candidate must have to pass the 1st exam.

So, $P\left(\frac{E_2}{E_1}\right)=0.8\;\cdots(ii)$

where $P(E_2/E_1)$ denotes the probability of passing the 2nd exam when she already passed the 1st exam (given, 0.8).

Similarly, as the conditional probability of passing the 3rd exam is 0.7, and the condition for this is, at first, she must have to pass the 1st and 2nd exam. i.e,

$P\left(\frac{E_2}{P(E_2/E_1)}\right)=0.7\;\cdots(iii)$

(a) For passing all the exams, the condition is, at first, she has to pass the 1st and 2nd exam, then she has to pass the 3rd exam too. The probability for this conditional has been given as 0.7.

So, the probability that she passes all three exams is 0.7.

(b) Given that she didn't pass all three exams that means she either failed in 1st exam or she passed the 1st and failed in 2nd exam or she passed both 1st and 2nd but failed in the 3rd exam.

Let F be the event that she didn't pass all three exams. So,

$P(F)=(1-P(E_1))+\left(1-\frac{P(E_2)}{P(E_1)}\right)+\left(1-\frac{P(E_2)}{P(E_2/E_1)}}\right)$