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3 years ago

How to solve this? I don't get it

Mathematics

1 answer:

Neporo4naja [7]3 years ago

8 0

You dont solve it. see the line at the bottom thats your line plot. number it from 1 through 8 and put the numbers where they belong on the line.

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How to find rate of change on a table

LekaFEV [45]

Answer:

__X__l__Y_

1 l 3

5 l 6

9 l 9

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21 l 18

On the right side, the numbers get larger by adding three each time. On the left side, the numbers get larger by adding 4 each time. It would be the same for any other pair of numbers when you add 4 to the left, and 3 to the right. For instance, the next pair would be (25 , 21). 25 would be the X value, and 21 would be the Y value.

4 0

4 years ago

When a store had Sold 2/5 of the shirts that were on display, they brought out another 30 from the stockroom. The store likes to

Inga [223]

Answer:

Part A) 3/5 is the total number of shirts remaining on display after selling 2/5

Part B) $x \geq 200$ (All real whole numbers greater than or equal to 200 shirts)

Part C) Initially on display were the amount of 200 or more shirts

Step-by-step explanation:

Let

x-------> total number of shirts on display

we know that

The inequality that represent the situation is

$(1-\frac{2}{5})x+30 \geq 150$

$(\frac{3}{5})x+30 \geq 150$

Part A) Explain what 3/5 means in the inequality

3/5 is the total number of shirts remaining on display after selling 2/5

Part B) Solve the inequality

we have

$(\frac{3}{5})x+30 \geq 150$

Subtract 30 both sides

$(\frac{3}{5})x \geq 150-30$

$(\frac{3}{5})x \geq 120$

Multiply by 5/3 both sides

$x \geq 120(5/3)$

$x \geq 200\ shirts\ on\ display$

Part C) Explain what the solution means in the situation

1) Initially on display were the amount of 200 or more shirts

2) 2/5 were sold and 120 or more shirts were left on display.

3) The store brought out another 30 shirts from the stockroom and placed them on display, for a total of 150 shirts or more on display

8 0

3 years ago

TheValue of a stock begins at $0.05/and increase by 0.01 each month write an equation representing the value of the stock v in a

Ne4ueva [31]

Because you start with $0.05, you're going to add that to whatever 0.01 is equal to after the amount of months

the equation I got was: v=0.05+(0.01m)

3 0

3 years ago

Given the polynomial expression 3x^2 + 3bx - 6x - 6b, factor completely.

My name is Ann [436]

The answer is 3 (x+b) (x-2)

Yw! :)

3 0

3 years ago

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers ( h, i , j , m ), are there with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

$\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]$

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

4 0

3 years ago