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Anni [7]
3 years ago
10

If two fair dice are rolled, what’s the probability that the product of the

Mathematics
1 answer:
Oksanka [162]3 years ago
8 0

Answer:

4/1. ( not so sure tho)

Step-by-step explanation:

=[(6/6)×(6/6)] + [(5/6)×(6/6)] + [(6/6)×(5/6)]

=(36/36) + (30/36) + (30/36)

=(taking LCM) 96/36

=4/1.

hope this helps

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Which equation has no solution?
vovangra [49]
C. -5 + 9 = 22 + 9 - 7 is the answer.

This is because when solved it’s : -52 = 15 which is false therefore there’s no solution.
6 0
3 years ago
What is the perimeter of rhombus LMNO?
kicyunya [14]

If you see that 3x-3 and set it equal to x+7 then you will get this,

3x-3=x+7


Now solve for X. You will get 5.

Then you plug five in and get 12 For both sides and then multiply 12 by 4 and get 48.

You answer is 48.

Hope this helps you.

4 0
3 years ago
Read 2 more answers
A coin is tossed and a single 6 sided number cube is rolled. Find the probability of landing on the heads side of the coin and r
Delicious77 [7]

Answer:

The probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

Step-by-step explanation:

Given:

Coin is tossed and a single 6 sided number cube is rolled.

Let the event of tossing head be 'H' and event of rolling a 3 be 'R3'.

Now, we know that:

Probability of an event 'A' = Favorable outcomes of 'A' ÷ Total possible outcomes

Now, for tossing a coin, the total possible outcomes are head and tail. So, there are 2 possible outcomes.

So, probability of event 'H' is given as:

P(H)=\dfrac{n(H)}{n(S)}\\\\P(H)=\frac{1}{2}=0.5

Similarly, for rolling a 6 sided cube, the possible outcomes are numbers 1 to 6. So, the number of possible outcomes is, n(S)=6

Now, probability of rolling a 3 is given as:

P(R3)=\frac{n(R3)}{n(S)}\\\\P(R3)=\frac{1}{6}

Now, both the events 'H' and 'R3' occur together. So, the combined probability is the product of two individual probabilities as they are independent events. So,

P(H\ and\ R3)=P(H)\times P(R3)\\\\P(H\ and\ R3)=\frac{1}{2}\times \frac{1}{6}\\\\P(H\ and\ R3)=\frac{1}{12}=0.083

Note: Independent events are those events whose intersection is an empty set of events or the outcome of one event doesn't affect the outcome of the other event.

Therefore, the probability of landing on the heads side of the coin and rolling 3 on the number cube is 0.083.

4 0
3 years ago
Julia is casting a play with 4 main roles. Six students
aev [14]

Using the combination formula, it is found that Julia can take 15 combinations.

<h3>What is the combination formula?</h3>

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by:

C_{n,x} = \frac{n!}{x!(n-x)!}

For this problem, 4 students are taken from a set of 6, hence the number of combinations is given as follows:

C_{6,4} = \frac{6!}{4!2!} = 15

More can be learned about the combination formula at brainly.com/question/25821700

#SPJ1

7 0
2 years ago
Someone please help me
Step2247 [10]

Answer:

20 sq in

Step-by-step explanation:

formula is (base * height)/2 so plug it in:

5*8=40

40/2=20

20 sq in

8 0
1 year ago
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