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Dvinal [7]
2 years ago
14

bella rolls 2 standard dice 60 times. how many times can she expect the sum of the numbers to be greater than 10?

Mathematics
1 answer:
iris [78.8K]2 years ago
6 0

Answer: in 5.

Step-by-step explanation:

The summ is greater when:

D1 = 5, D2 = 6

D1 = 6, D2 = 5

D1 = 6, D2 = 6

3 outcomes out of the 36 possible outcomes (6 for each dice, 6*6 = 36) for rolling 2 dice,

The relative probability is:

p = 3/36

Then, in 60 rolls, we can expect that:

(3/36)*60 = 5

In 5 rolls the sum of the numbers is greater than 10.

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Multiply a number n by 5, and then add 3 to product
joja [24]

Answer:

5n+3

Step-by-step explanation:

4 0
3 years ago
Solve the equation : 3x - 2 = 4x + 5<br>How many solutions are there?​
Nookie1986 [14]

Answer:

one solution

Step-by-step explanation:

Given

3x - 2 = 4x + 5 ( subtract 4x from both sides )

- x - 2 = 5 ( add 2 to both sides )

- x = 7 ( multiply both sides by - 1 )

x = - 7 ← the one solution

8 0
3 years ago
Read 2 more answers
Find the amount that results from the given investments. $200 invested at 9% compounded after a period of 4 years
Yanka [14]

Answer:

The amount of invested money after a period of 4 years:

A = 200 x (1 + 9/100)^4 = 282.32 dollar

Hope this helps!

:)

8 0
2 years ago
Find the measures of the angles of the triangle whose vertices are A = (-3,0) , B = (1,3) , and C = (1,-3).A.) The measure of ∠A
alekssr [168]

Answer:

\theta_{CAB}=128.316

\theta_{ABC}=25.842

\theta_{BCA}=25.842

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

AB= \sqrt{(-3-1)^2+(0-3)^2}=5

BC= \sqrt{(1-1)^2+(3-(-3))^2}=9

CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

the angle BCA is opposite to the AB.

to find the CAB, we'll use:

BC^2 = AB^2+CA^2-(AB)(CA)\cos{\theta_{CAB}}

\dfrac{BC^2-(AB^2+CA^2)}{-2(AB)(CA)} =\cos{\theta_{CAB}}

\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}

\theta_{CAB}=\arccos{-\dfrac{0.62}}

\theta_{CAB}=128.316

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}

\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)

\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)

\theta_{ABC}=\arcsin{0.4359}\right)

\theta_{ABC}=25.842

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

\theta_{BCA}=25.842

this can also be checked using the fact the sum of all angles inside a triangle is 180

\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180

25.842+128.316+25.842

180

6 0
3 years ago
Read 2 more answers
A private jet flies the same distance in 8 hours that a commercial jet flies in 7 hours. If the speed of the commercial jet was
mamaluj [8]

Solving a system of equations, we can see that:

  • Speed of the private jet: 168 mi/h
  • Speed of the commercial jet: 192mi/h

<h3>How to find the speeds of each jet?</h3>

Let's define the variables:

  • P = speed of the private jet.
  • C = speed of the commercial jet.

With the given information, we can write:

P*8h = D

C*7h = D

C = 2*P - 144mi/h

So we have a system of 3 equations, where D is the distance in the problem.

With the first and second equations we can write:

P*8h = D = C*7h

Isolating P, we get:

P = C*(7/8)

Now we can replace that in the last equation:

C = 2*P - 144mi/h

C = 2*C*(7/8) - 144mi/h

And now we can solve that for C.

C - 2*(7/8)*C = - 144mi/h

C*(1 - 14/8) = -144mi/h

C*(8/8 - 14/8) = - 144mi/h

C*(6/8) = 144mi/h

C = (8/6)*144mi/h = 192mi/h

Now that we know the speed of the commercial jet, we can find the speed of the private jet.

P = C*(7/8) = 192mi/h*(7/8) = 168 mi/h

If you want to learn more about systems of equations:

brainly.com/question/13729904

#SPJ1

6 0
2 years ago
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