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ASHA 777 [7]
2 years ago
5

A dog sleep from 10:36 PM to 3:36 AM.How long did the dog sleep?

Mathematics
2 answers:
Sladkaya [172]2 years ago
5 0

Answer:

<h2>5 hours</h2>

Step-by-step explanation:

10:36- 11:00=24 min

11:00-3:00= 4 h

3:00-3:36=36 min

add them up

=

5h

jolli1 [7]2 years ago
4 0

Answer:

5 hours.

Step-by-step explanation:

Note that the clock "restarts" at 12, and flips from PM to AM (and vice versa).

In this case:

1st hour: 10:36 - 11:36pm

2nd hour: 11:36pm - 12:36am

3rd hour: 12:36am - 1:36am

4th hour: 1:36am - 2:36am

5th hour: 2:36am - 3:36am

~

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Someone help me with this
Len [333]

Answer:

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4 0
2 years ago
A) How many ways can 2 integers from 1,2,...,100 be selected
Anna007 [38]

Answer with explanation:

→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

7 0
2 years ago
Can somebody who knows how to do probability please help answer these questions correctly? Thanks! (Btw the P= probability)
stira [4]

Answer:

P(2): 1/5

P(4): 1/5

P(odd number): 3/5

P(whole number): 5/5

P(6): 0/5

P(2 or 3): 2/5

Step-by-step explanation:

There are 5 <em>equal </em>sections in this circle. So, the probability to land in each section is 1/5.

The odd numbers are 1, 3, and 5. Since each section is 1/5, you add

1/5 + 1/5 + 1/5 = 3/5. That is the probability that you will land in any odd number.

Because all the numbers listed are whole numbers, no matter where the spinner lands it will be a whole number. So, the probability is 5/5 for whole numbers.

Since 6 is not a section, it's probability will be 0/5 (or you can just put 0).

"Or" means you add the two probabilities. Add the probability of landing on 3 (which is 1/5) to the probability of landing on 2 (which is also 1/5). So, you get 2/5.

6 0
3 years ago
Read 2 more answers
Question 1 options:<br><br> 13.5<br><br><br> 12<br><br><br> 13<br><br><br> 11
ki77a [65]

Answer: 13

nothing in between 13

Down below

4 0
2 years ago
Please help! Brainliest if you explain
My name is Ann [436]
Switch sides

4/5 (x+3)=y+6

Multiply both sides by 5

5*4/5 (x+3) = 5y+5*6

Simplify

4(x+3) = 57+30

Divide both sides by 4

4(x+3)/4 = 5y/4+ 30/4

Simplify

x+3 = 5y+30/4 

Subtract 3 from both sides

x+3 - 3 = 5y+30/4 - 3

Simplify 

x=5y+30/4 - 3











3 0
3 years ago
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