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seropon [69]
3 years ago
13

Write a numerical expression that has a value equal to each number from 1 through 30. In each expression, use each of four conse

cutive digits from 0–9 exactly once.
Mathematics
1 answer:
Hunter-Best [27]3 years ago
5 0

Answer:

See Explanation

Step-by-step explanation:

Required

Use any 4 consecutive digit from 0 to 9 to form an expression that equals 1 through 30

There is no formula to answer this question. All we need to do is, apply trial by error method to for each expression.

Possible expression and their results are:

- 1  + 2 * 3 - 4 = 1

1 + 2 + 3 - 4 = 2

0 * 3 + 1 + 2 = 3

0 * 2 + 1 + 3 = 4

1 * 3 + 4 - 2 =5

0 + 1 + 2 + 3 = 6

0 + 1 + 2 * 3 = 7

3 + 4 - 5 + 6 = 8

1 * 2 + 3 + 4 =9

1 + 2 + 3 + 4 =10

2^3 + 4 - 1 = 11

5 + 6 - 7 + 8  =12

4^2 - 3 * 1 = 13

2 + 3 + 4 +5=14

2 * 3 +4 + 5 = 15

4 * (3 - 1)^2 = 16

3 * 5 + 4/2 = 17

3 + 4 + 5 + 6 = 18

2 + 3 * 4 + 5 = 19

3 + 2^4 + 1 = 20

4 * 5 - 6 + 7 = 21

4 + 5 + 6 + 7 =22

3 * 4 + 5 + 6 = 23

5^2 -4 + 3 = 24

2^5 - 4 - 3 = 25

5 + 6 + 7 +8 = 26

\frac{3^4}{5-2} =27

5 * 4 + 2^3 = 28

3 + 4 * 5 + 6 = 29

6 + 7 + 8 +9 = 30

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Solving a Direct Variation Problem

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

Second one

third one

fifth one

last one

Step-by-step explanation:

Based on the x intercepts we can write

a(x+1)(x-5)

where a is some constant

solve for a by plugging in some coordinates

let's plug in (1,-8)

-8=a(1+1)(1-5)

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a=1

therefore the quadratic is (x+1)(x-5)

expand this

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To find out where something is decreasing/increasing it's easiest to take the first derivative

x²-4x-5= 2x-4

set this equal to 0

2x-4=0

2x=4

x=2

Which means that our two intervals are

(-∞,2)U(2,∞)

plug in any values in the intervals to see whether or not the function is increasing/decreation

Let's plug in 0 for (-∞,2)

2(0)-4

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It's negative so it's decreasing on this interval

DO the same thing with the other one

let's plug in x=3

2(3)-4

2

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Go back to the critical value of x=2 and plug this into the equation to find the max/min

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6 0
3 years ago
The compressive strength of concrete is normally distributed with mu = 2500 psi and sigma = 50 psi. A random sample of n = 8 spe
Nata [24]

Answer:

X \sim N(2500,50)  

Where \mu=2500 and \sigma=50

We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error would be:

SE = \frac{50}{\sqrt{8}}= 17.678 psi

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Let X the random variable that represent the compressive strength of concrete of a population, and for this case we know the distribution for X is given by:

X \sim N(2500,50)  

Where \mu=2500 and \sigma=50

We select a sample size of n =8. Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the standard error would be:

SE = \frac{50}{\sqrt{8}}= 17.678 psi

7 0
3 years ago
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