The average of any two consecutive even numbers in an even number. Find a counterexample.

Mathematics

1 answer:

pishuonlain [190]3 years ago

6 0

Answer:

2 and 4

Step-by-step explanation:

2+4 = 6

6/2 = 3

average is sum of number divided by how many numbers there are.

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In a recent year, 31% of all college students were enrolled part-time. If 8.6 million college students were enrolled part-time t

Ronch [10]

Answer:

28 million college students.

Step-by-step explanation:

We can represent all college students as x. This means that all part-time college students are $\frac{31}{100}x$ . Because all part-time college students are also equal to 8.6 million, we have $8,600,000=\frac{31}{100}x$ . To find the number of total college students, or x, we have to multiply by 100/31 to get rid of the fraction on the other side:

$8,600,00\cdot\frac{100}{31}=x$

The left side of the equation evaluates as $27,741,935.5$ . Because we are asked to round to the nearest million, we have to round to 28 million, which is the closest.

5 0

2 years ago

I have 1 hundreds, 4 ones, 2 tens, and 1 tenths. What number am I?

DaniilM [7]

Since it reads 1 hundreds, that means 100. 4 ones mean 4 * 1, which is 4. 2 tens means 2 * 10, which is 20. Add those together and you get 124. Where it reads "1 tenth", that means 0.1
Therefore, the number is 124.1

6 0

4 years ago

single die is rolled twice. Find the probability of rolling an oddodd number the first time and a number greater than 33 the sec

givi [52]

Answer:

$\frac{1}{4}$

Step-by-step explanation:

A single die is rolled twice, we have to find the probability of rolling an odd number in first throw and a number greater than 3 in the second throw.

a) Rolling an Odd number in first throw

A die has total 6 possible outcomes, out of which 3 are odd numbers i.e. 1,3 and 5

So, total number of possible outcomes = 6

Total Favorable outcomes (Odd numbers) = 3

Probability is defined as the ratio of favorable outcomes to total number of outcomes. So,

The probability of rolling an odd number would be = $\frac{3}{6} = \frac{1}{2}$

b) Rolling a number greater than 3 in second throw

Here again total possible outcomes = 6

Favorable outcomes (Numbers greater than 3 are 4, 5 and 6) = 3

So,

The probability of rolling a number greater than 3 = $\frac{3}{6} = \frac{1}{2}$

These two events(rolls) are independent of each other, so the overall probability of both events occurring would be the product of individual probabilities.

So,

Probability of rolling an odd number the first time and a number greater than 3 the second time = $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$