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3 years ago

1 11/15 + 6 1/6 =what is the answer to this problem?

Mathematics

2 answers:

Anna71 [15]3 years ago

6 0

Answer:7 27/30

Step-by-step explanation

So yeah

kakasveta [241]3 years ago

6 0

Answer:

Hi! The answer is 7 27/30

Step-by-step explanation:

You might be interested in

allsm [11]

Answer:

8:27

Step-by-step explanation:

Let Small cone be A, Large cone be B.

#Compare volumes to find scale factor.

$\ Volume \ scale \ factor= \frac{Vol A}{Vol B}\\=\frac{320}{1080}\\=\frac{8}{27} \ or\ 8:27$

5 0

3 years ago

Lina20 [59]

The probability of not drawing a black marble is 4/5

The probability of drawing a red marble is 3/10

<h3>What are the probabilities?</h3>

Probability determines the odds of the occurrence of a random event. The probability the event occurs is 1 and the probability that the event does not occur is 0.

Probability of not drawing a black marble = total marbles that are not black / total number of marbles

(5 + 3) / (5 + 3 + 2) = 8/10 = 4/5

The probability of drawing a red marble = total number of red marbles / total number of marbles

= 3/10

To learn more about probability, please check: brainly.com/question/13234031

#SPJ1

3 0

2 years ago

Which of the following numbers has an odd number of factors 13,15,25,26?

kvv77 [185]

Answer:

25. Its cubed.

Step-by-step explanation:

7 0

4 years ago

A dormitory has n students, all of whom like to gossip. one of the students hears a rumor, and tells it to one of the other n −

Alekssandra [29.7K]

Answer:

$P(r)=\frac{(n-3)(n-4)....(n-r)}{(n-2)^{r-2}}$

Step-by-step explanation:

From the question, we have the following condition:

$p_1=p_2=1,\:and\:p_n=0$

We know that each student who hears the rumor tells it to a student picked at random from the dormitory (excluding, of course, himself/herself and the person from whom he/she heard the rumor)

The 3rd student can therefore tell the rumour to n-2 students but only n-3 will accept it.

$\implies p(3)=\frac{n-3}{n-2}$

Consequently, the 4th student must not tell the 1st and second students.

$\implies p(4)=\frac{n-3}{n-2}\times \frac{n-4}{n-2}$

We can rewrite this to observe a pattern:

$\implies p(4)=\frac{(n-3)(n-4)}{(n-2)^2}$

$\implies p(4)=\frac{(n-3)(n-4)}{(n-2)^{4-2}}$

$\implies p(r)=\frac{(n-3)(n-4)(n-5)...(n-r)}{(n-2)^{r-2}}$

Hence, the probability that the rumor is told r times without coming back to a student who has already is:

$P(r)=\frac{(n-3)(n-4)....(n-r)}{(n-2)^{r-2}}$

See attachment for complete question

4 0

4 years ago

Adult tickets to a play cost $22. Tickets for children cost $15. Tickets for a group of 11 people cost a total of $228. Write an

yaroslaw [1]

Adults = 9
Children = 2

Let x be the number of adults.
11-x is the number of children.

22x + 15(11-x) = 228
22x + 165 - 15x = 228
22x - 15x = 228 - 165
7x = 63
7x / 7 = 63 / 7
x = 9 number of adults.

11 - x = 11 - 9 = 2 number of children.

To check:

22x + 15(11-x) = 228
22(9) + 15(11-9) = 228
198 + 30 = 228
228 = 228

7 0

4 years ago

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