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Pachacha [2.7K]
3 years ago
12

How to find a perfect square

Mathematics
1 answer:
Rzqust [24]3 years ago
8 0

Answer:

Step-by-step explanation:

The square of a number n is denoted by n^2.

Examples:  The square of 2 is 2^2, or 4; the square of 5 is 5^2, or 25.  To find a perfect square of a number n, multiply the number by itself (obtaining the square of n).

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Which one is greater 3/10 or 2/5
Andreyy89

Answer:

2/5

Step-by-step explanation:

2/5=4/10

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3 years ago
WILL MARK FIRST CORRECT ANSWER BRAINLIEST
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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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