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schepotkina [342]
3 years ago
9

How many integers are there that consist of four digits (from 1000 to 9999) where the first digit is a 5 or an 8 and the third d

igit is not 5 or 8
Mathematics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer:

1600 integers

Step-by-step explanation:

Since we have a four digit number, there are four digit placements.

For the first digit, since there can either be a 5 or an 8, we have the arrangement as ²P₁ = 2 ways.

For the second digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

For the third digit, since it neither be a 5 or an 8, we have two less digit from the total of ten digits which is 10 - 2 = 8. So, the number of ways of arranging that is ⁸P₁ = 8.

For the last digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

So, the number of integers that can be formed are 2 × 10 × 8 × 10 = 20 × 80 = 1600 integers

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nataly862011 [7]

Answer:

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3 years ago
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Mandarinka [93]
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Read 2 more answers
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SSSSS [86.1K]

Answer:

A) (17 ; 550)

B) $17/item

C) 550

Step-by-step explanation:

First we must calculate the intersection point of the two lines. Since in that point <em>y</em> has the same value in both equations, we can obtain <em>x </em>by equalling the two equations and then using that value for obtaining <em>y</em>:

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Supply and demand are in equilibrium when the amount of items on supply are the same as the ones on demand. That is the point were the two lines intersect, which means the selling price is the <em>x</em> coordinate and the amount of items is the <em>y</em> coordinate, so that is a selling price of <em>$17/item</em> with a number of items of <em>550</em>.

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Step-by-step explanation:

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3 years ago
Can Someone Please Explain This To Me
Ghella [55]

Answer:

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Step-by-step explanation:

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