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Mashutka [201]
2 years ago
11

" \frac{5}{3} + \frac{4}{6} = " alt=" \frac{5}{3} + \frac{4}{6} = " align="absmiddle" class="latex-formula">
.................​
Mathematics
1 answer:
o-na [289]2 years ago
3 0

Answer:

\frac{7}{3} =2\frac{1}{3}

Step-by-step explanation:

For the problem:

\frac{5}{3} +\frac{4}{6} = ?

The Least Common Multiple (LCM) of 3 and 6 is 6. Multiply the numerator and denominator of each fraction by whatever value will result in the denominator of each fraction being equal to the LCM:

\frac{5}{3} +\frac{4}{6} =\frac{5*2}{3*2} +\frac{4}{6} =\frac{10}{6} +\frac{4}{6}

Now that the fractions have like denominators, add the numerators:

\frac{10}{6} +\frac{4}{6} =\frac{10+4}{6} =\frac{14}{6}

The Greatest Common Factor (GCF) of 14 and 6 is 2. The result can be simplified by dividing both the numerator and denominator by 2.

\frac{14}{6} =\frac{14/2}{6/2} =\frac{7}{3}

7 ÷ 3 = 2R1, therefore:

\frac{7}{3} =2\frac{1}{3}

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3 0
3 years ago
The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes
Artist 52 [7]

Answer:

a) 0.023

b) 0.286

c) 10 students will be unable to complete the exam inthe allotted time.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 80 minutes

Standard Deviation, σ = 10 minutes

We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(completing the exam in one hour or less)

P(x < 60)

P( x < 60) = P( z < \displaystyle\frac{60 - 80}{10}) = P(z < -2)

Calculation the value from standard normal z table, we have,  

P(x < 60) =0.023= 2.3\%

b) P(complete the exam in more than 60 minutes but less than 75 minutes)

P(60 \leq x \leq 75) = P(\displaystyle\frac{60 - 80}{10} \leq z \leq \displaystyle\frac{75-80}{10}) = P(-2 \leq z \leq -0.5)\\\\= P(z \leq -0.5) - P(z < -2)\\= 0.309- 0.023 = 0.286= 28.6\%

c) P(completing the exam in more than 90 minutes)

P(x > 90)

P( x > 90) = P( z > \displaystyle\frac{90 -80}{10}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 90) = 1 - 0.8413 = 0.1587 = 15.87\%

15.87% of children of class will require more than 90 minutes to complete the test.

Number of children =

\dfrac{15.87}{100}\times 60 = 9.52\approx 10

Approximately, 10 students of class will require more than 90 minutes to complete the test.

4 0
3 years ago
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