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3 years ago

You roll 2 dice and flip a coin. What's the probability of getting two 4's and a tails?

Mathematics

1 answer:

zubka84 [21]3 years ago

7 0

Answer:

2/4

Step-by-step explanation:

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Is the question ''How many cars were sold each day this month'' statistical or not?

kifflom [539]

2,000 cars............

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4 years ago

Why does the fraction 20/25 not belong with 1/20, 2/3 and 5/4?

devlian [24]

I’m not 100% sure but I think it’s because 20/25 can still be simplified further, while the other 3 can’t.

7 0

3 years ago

8 7 / 8 + 9 13 / 16 what is this in a mixed number.

FinnZ [79.3K]

Answer:

18 and 11/16

Step-by-step explanation:

I converted the two numbers into improper fractions which leaves me with

71/8+157/16

I thenn made sure I had common denominators, which means I need to multiply the first number by 2/2 and I'm left with:

142/16+157/16

I can now add the 2 numbers together and I get:

299/16

now we just need to convert the number into a mixed number by dividing, and I am left with:

18 and 11/16

5 0

3 years ago

A subway train makes six stops of equal length during its 21 km run. The train is actually moving for 20 minutes of the trip. At

photoshop1234 [79]

Answer:

The subway train can stop for a maximum of 2.5 minutes on each stop.

Step-by-step explanation:

3 0

3 years ago

A box contains 35 gems, of which 10 are real diamonds and 25 are fake diamonds. Gems are randomly taken out of the box, one at a

USPshnik [31]

Answer:

$Probability = 0.504$

Step-by-step explanation:

Given

$Gems= 35$

$Real = 10$

$Fake = 25$

Required

Determine the probability of selecting two fakes

The probability can be represented as thus: $P(Fake\ and\ Fake)$

Using the following probability formula, we have:

$P(Fake\ and\ Fake) = P(Fake) * P(Fake)$

Each probability is calculated by dividing number of fakes by total number of gems:

$P(Fake\ and\ Fake) = \frac{25}{35} * \frac{25-1}{35-1}$

The minus 1 (-1) represent the numbers of fake and total gems left after the first selection

$P(Fake\ and\ Fake) = \frac{25}{35} * \frac{24}{34}$

$P(Fake\ and\ Fake) = \frac{5}{7} * \frac{12}{17}$

$P(Fake\ and\ Fake) = \frac{60}{119}$

$P(Fake\ and\ Fake) = 0.504$

Hence, the required probability is approximately 0.504

7 0

3 years ago