Oliver can read 28 1/2 pages in 1 1/6 hours. how long does it take him to read one page

If two standard six-sided dice are tossed, what is the probability that a 5 is rolled on at least one of the two dice? express y

zaharov [31]

We define the probability of a particular event occurring as:
$\frac{number\ of \ desired\ outcomes}{number\ of\ possible\ outcomes}$

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are independent, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for each of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.

Now that we've found the number of possible outcomes, we need to find the number of desired outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is at least one 5 rolled. It turns out, there are only 3:

(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5

So, we have $\frac{3}{36} = \frac{1}{12}$ probability of rolling at least one 5.

6 0

3 years ago

Find the area of circle with a radius of 2 in exact terms

ss7ja [257]

Answer:

12.56 (13 if rounded)

Step-by-step explanation:

A=pi x r squared

A=2 squared multiplied by pi

A=4 times pi

plug into calculator should be about 12.56

7 0

3 years ago

I couldn’t find the answer throught brainly. Anyone know??

STatiana [176]

Answer:

240 would be your answer

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

3 years ago

PLEASE HELP! WILL MARK BRAINLIEST!! :D

netineya [11]

Hope this helps

Given a term in a geometric sequence and the common ratio find the first five terms, the explicit formula, and the recursive formula. Given two terms in a geometric sequence find the 8th term and the recursive formula. Determine if the sequence is geometric. If it is, find the common ratio.

6 0

2 years ago