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maria [59]
3 years ago
5

1. You are rolling a pair of dice. What's the probability of getting different numbers on both dice?

Mathematics
1 answer:
GaryK [48]3 years ago
7 0

Answer:

  • 5/6

Step-by-step explanation:

<u>Total outcomes:</u>

  • 6*6 = 36

<u>Outcomes with same numbers on both dice:</u>

  • 6

<u>Outcomes with different numbers on both dice:</u>

  • 36 - 6 = 30

<u>Probability of different numbers on both dice:</u>

  • 30/36 = 5/6
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A crew of highway workers paved 1/5 mile in 1/8 hour. If they work at the same rate, how much will they
Elan Coil [88]

Answer:

The easiest approach is to realise that one hour is 3 times longer than 20 minutes. The longer the time, the more they will pave.

215 of a mile, in 20 minutes, how much in 60 minutes?#

They will pave 3 times more.

215×31=615 of a mile

615=25 of a mile

You could also use the 'unitary method' where you find out how much they pave in ONE minute (divide by 20) and them multiply by 60 to find how much in one hour.

Look at what happens:

215÷20×60

=215×120×603

=215×3 ← exactly the same maths.

=25

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3 years ago
Angie and Kim are sharing a large sub sandwich. Angie ate
VikaD [51]

They ate 2/3 together. And, they have 1/3 left!

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Ac is tangent to circle o at point c what is the measure of angle o?<br>​
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Assuming this triangle is a right triangle, 52 degrees.
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What is the product of seven plus two and eight minus two
mr_godi [17]

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Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline
vivado [14]

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 z = \frac{x-np}{\sqrt{np(1-p)}  } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} } =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = \sqrt{npq} = 2.233

P (x≤100) = P(z = P(z<0.34) = 0.63307.

6 0
3 years ago
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