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3 years ago

How do you create a table of values

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

5 0

Basically set up two rows divided into however many section you need. Row 1 is the x values, Row 2 is the y values

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The sum of 5 consecutive integers is 505

kramer

Using algebra:n+(n+1)+(n+2)+(n+3)+(n+4)=5055n+10=5055n=505-105n=495n=495/5n=99n+1=100n+2=101 answer

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3 years ago

10,000,000 in scientific notation

san4es73 [151]

The answer is 1 x 10^7

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3 years ago

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after an oil pipeline burst one morning gas prices went up by $2.20 per gallon at the afternoon you bought 10 gallons of gas for

mr Goodwill [35]

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3 years ago

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Simply the following expressions:

Lapatulllka [165]

Answer:

6.  (2x - 1) 

(3x + 2)

8.  y - 3 

y² - 3y + 9

Step-by-step explanation:

6. 2x² + 9x - 5

3x² + 17x + 10

= (2x - 1) (x + 5)

3x² + 17x + 10

= (2x - 1) (x + 5)

(3x + 2) (x + 5)

=  (2x - 1) 

(3x + 2)

8. y² - 9

y³ + 27

= (y + 3) (y - 3)

y³ + 27

= (y + 3) (y - 3)

(y+3) (y² - 3y + 9)

=  y - 3 

y² - 3y + 9

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3 years ago

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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

4 years ago