MATHEMATICS MIDDLE SCHOOL

Two fair dice are tossed.a. What is the probability that the sum of the number of dots shown on the upper faces is equal to 7? To 11?
b.What is the probability that you roll "doubles" that is, both dice have the same number on the upper face?
c. What is the probability that both dice show an odd number?

Answers

Answer 1

Answer:

there are 6 possibilities on each dice. Therefore 6 x 6 = 36 possible outcomes.

What is the probability that the sum of the number of dots shown on the upper faces is equal to 7?

1,6 2,5 3,4 4,3 5,2 6,1 there are 6 possible ways to get 7

6/36 = 1/6 = .1666...

What is the probability that the sum of the number of dots shown on the upper faces is equal to 11?

5,6 6,5 there are 2 possible ways to get 11

2/36 = 1/18 = .0555...

What is the probability that you roll "doubles" that is, both dice have the same number on the upper face?

1,1 2,2 3,3 4,4 5,5 6,6 there are 6 possible ways to get doubles

6/36 = 1/6 = .1666...

What is the probability that both dice show an odd number?

1,1 1,3 1,5 3,1 3,3 3,5 5,1 5,3 5,5 there are 9 possible ways that both dice show an odd number

9/36 = 1/4 = .25

Answer 2

Answer:

Final answer:

The probability of getting a sum of 7 when two fair dice are tossed is 1/6. The probability of getting a sum of 11 is 1/36. The probability of rolling doubles is 1/6. The probability of both dice showing an odd number is 1/4.

Explanation:

a. Probability of sum equal to 7:

When two fair dice are tossed, there are a total of 36 possible outcomes, as each die has 6 possible outcomes. Out of these, there are 6 outcomes that result in a sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1).

Therefore, the probability of getting a sum of 7 is 6/36 = 1/6.

Probability of sum equal to 11:

There is only one outcome that results in a sum of 11, which is (5,6) or (6,5). Therefore, the probability of getting a sum of 11 is 1/36.

b. Probability of rolling doubles:

When rolling two dice, there are 6 possible outcomes that result in doubles: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). Since there are a total of 36 possible outcomes, the probability of rolling doubles is 6/36 = 1/6.

c. Probability of both dice showing an odd number:

Since each die has 3 odd numbers (1, 3, and 5), the probability of both dice showing odd numbers is (3/6) * (3/6) = 9/36 = 1/4.

Learn more about Probability and Statistics here:

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What is the algebraic expression to 6 less than twice a number

What are two decimals whose product is close to 10

Answers

There are millions of possible answers.

Here's one: 2.14 and 4.673 .

   2.14 x 4.673 = 10.00022

Another one: 2.55 and 3.922

   2.55 x 3.922 = 10.0011

Just one more: 7.5 and 1.3333

   7.5 x 1.3333 = 9.99975

Answer: 0.90 and 0.900

Hope this helped

I need help please I don't know if it is right my answer is x=16.66

Answers

I personally think it is right.

That answer is correct!

What is the prime factorization of 27

Answers

3x3x3
that is the prime

27 = 1 · 3 · 3 · 3

That's all there is to it.

A bus goes from town A to B in an exact time. If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to and if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier. Find:A) The distance between the two towns;
B) The exact time that it takes to arrive town B
C) The speed of the bus(by schedule) for the exact time.

Answers

Let the speed of bus for the exact time = x km/h
the distance between the cities = y km
then the exact time would be, t hours = (y/x) hours

If the bus goes at the rate of 50km/h, then it will arrive B 42min later,
speed = 50 km/h
42 minutes = 42/60 hours = 7/10 hours = 0.7 hours
time taken = t+0.7
distance = speed×time
⇒ y = 50×(t+0.7)
⇒ y = 50t + 35 ---------------------(1)

it increases its speed 5.5/9 m/sec, it will arrive B 30min earlier.
5.5/9 m/s = (5.5/9)×(18/5) km/h = 2.2 km/h
30 minutes = 30/60 = 0.5 hour
speed = (x+2.2) km/h
time = (t - 0.5) hours

distance = speed×time
⇒ y = (x+2.2)×(t-0.5)
⇒ y = ((y/t) +2.2)×(t-0.5) (t = y/x)
⇒ y = y - 0.5 (y/t) + 2.2t - 1.1
⇒ 0.5 (y/t) - 2.2t + 1.1 = 0   (subtracting y from both sides)
⇒ (y/t) - 4.4t - 2.2 = 0    (dividing both sides by 0.5)
⇒ y - 4.4t² - 2.2t = 0    (multiplying both sides by t)
⇒ 50t + 35 - 4.4t² - 2.2 t = 0 (from equation 1)
⇒ -4.4t² + 35 + 47.8t = 0
⇒ 4.4t² - 47.8t - 35 = 0

solving the quadratic equation, we get t = 11.55 hours
y = 50t + 35 = 612.5 km
x = 612.5/11.55 = 53 km/h

A) 612.5 km
B) 11.55 hours
C) 53 km/h

The Logic Defined:

1 Minute=t, (a unit of time)

Time (By schedule)=nt, (n>0), nt=number of minutes

Metre(s)=m

Speed=s (in metres per minute), s=[distance in metres]/[time in minutes]

Distance=d (in metres), d=[speed in metres per minute]*[time in minutes]

---------------------------------------------

Statement (1):

"If the bus goes at the rate of 50km/h, then it will arrive B 42min later than it had to."

Conclusion 1:

$\frac { 50km }{ h } =\frac { 50,000m }{ 60t } =\frac { 2,500m }{ 3t } \n \n \therefore \quad \frac { 2,500m }{ 3t } =nt+42t\n \n \frac { 2,500m }{ 3t } =t\left( n+42 \right)$

$\n \n 2,500m=3{ t }^( 2 )\left( n+42 \right) \n \n m=\frac { 3{ t }^( 2 )\left( n+42 \right) }{ 2,500 }$

Statement (2):

"if it increases its speed 5.5/9 m/sec., it will arrive B 30min earlier."

Conclusion 2:

$\frac { 5.5m }{ 9\quad seconds } =\frac { 5.5m }{ \frac { 9 }{ 60 } t } =\frac { 110m }{ 3t }$

$\n \n \therefore \quad \frac { 110m }{ 3t } =nt-30t\n \n \frac { 110m }{ 3t } =t\left( n-30 \right) \n \n 110m=3{ t }^( 2 )\left( n-30 \right)$

$\n \n m=\frac { 3{ t }^( 2 )\left( n-30 \right) }{ 110 }$

Conclusion 3, because of conclusion 1 and 2:

$\frac { 3{ t }^( 2 )\left( n-30 \right) }{ 110 } =\frac { 3{ t }^( 2 )\left( n+42 \right) }{ 2,500 } \n \n 7,500{ t }^( 2 )\left( n-30 \right) =330{ t }^( 2 )\left( n+42 \right) \n \n 7,500\left( n-30 \right) =330\left( n+42 \right)$

$\n \n 7,500n-225,000=330n+13,860\n \n 7,500n-330n=13,860+225,000\n \n 7,170n=238,860\n \n n=\frac { 238,860 }{ 7,170 } \n \n \therefore \quad n=\frac { 7962 }{ 239 }$

Therefore,

$Time\quad by\quad schedule=\frac { 7962 }{ 239 } t\n \n Approx:\quad 33.3\quad mins$

Now we want to find the distance between the two towns, so we say that:

$d=\frac { 2,500m }{ 3t } \cdot \left( \frac { 7962 }{ 239 } t+42t \right) \n \n =\frac { 2,500m }{ 3t } \cdot \frac { 18,000 }{ 239 } t$

$\n \n =\frac { 45,000,000 }{ 717 } m\n \n Approx:\quad 62,761.5\quad metres\n \n In\quad km\quad (approx):\quad 62.761\quad km$

So now you want to know how fast the bus has to travel to get to its destination on time...

Use the formula: s=d/t

Therefore:

$s=\frac { \frac { 45,000,000 }{ 717 } m }{ \frac { 7962 }{ 239 } t } \n \n Approx:\quad 1,883.9\quad metres\quad per\quad minute$

A park has t tables spread equally among the 3 picnic areas. Write an expression that shows how many picnic tables are at each picnic area.

Answers

Answer:

Number of picnic areas =3

Step-by-step explanation:

Answer:

t/3

Step-by-step explanation:

add all the tables in each picnic areas, since the tables are spread equally among 3 areas. Divide the number of table by 3 to get the number of tables at each picnic area.

so, all tables divided by 3. Since we do not know how many tables are there, let put t = numbers of table

t ÷ 3

t/3

Mr. Hogan's class has 16 students in it. On picture day 5 students stand in the first row, 4 students stand in the second row, 3 students stand in the third row, and 4 students stand in the fourth row. If a random student blinks what is the probability that the student will be in the second or third row?A. 1/4
B. 3/16
C. 5/16
D. 7/16

Answers

d. 7/16
i think its that because 4 plus 3 equals 7 and they asked second or third row and they only have second row up there and they asked for both so i think 7/16.

The answer is D.7/16 due to the fact that there are 4 students in the second row, and 3 students in the third, the ratio would be 7/16.hope i helped.

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