All categories

3 years ago

How how many students hade more than 6 candy bars (use the picture above)

Mathematics

2 answers:

Cloud [144]3 years ago

4 0

Answer:

About 5 if your are including 6, or just 3.

Step-by-step explanation:

To find it you can just count the numbers greater than 6.

Hope this helps.

brainliest if correct

aleksandr82 [10.1K]3 years ago

3 0

Answer:

3 Students

Step-by-step explanation:

There is only two more slots for how many candy bars each student has, in the "7" slot there is only one student that has seven candy bars and in the "8" slot there is two students that had eight candy bars.

1+2=3

I hope that helped.

You might be interested in

The price of an article is increased from Rs 75 to Rs 78. Find the percentage increase in the price of the article.

andreev551 [17]

( 78 -75 / 75 ) × 100 =

(3/75 )× 100 =

( 3 / 3 × 25 ) × 100 =

(1/25) × 100 =

100/25 =

% 4

4 0

3 years ago

How many solutions does the system of equations below have?

soldier1979 [14.2K]

Answer:

One solution

Step-by-step explanation:

5x + y = 8

15x + 15y = 14

Lets solve using substitution, first we need to turn "5x + = 8" into "y = mx + b" or slope - intercept form

So we solve for "y" in the equation "5x + y = 8"

5x + y = 8

Step 1: Subtract 5x from both sides.

5x + y − 5x = 8 − 5x

Step 2: 5x subtracted by 5x cancel out and "8 - 5x" are flipped

y = −5x + 8

Now we can solve using substitution:

We substitute "-5x + 8" into the equation "15x + 15y = 14" for y

So it would look like this:

15x + 15(-5x + 8) = 14

Now we just solve for x

15x + (15)(−5x) + (15)(8) = 14(Distribute)

15x − 75x + 120 = 14

(15x − 75x) + (120) = 14(Combine Like Terms)

−60x + 120 = 14

Step 2: Subtract 120 from both sides.

−60x + 120 − 120 = 14 − 120

−60x = −106

Divide both sides by -60

$\dfrac{ -60x }{ -60 } = \dfrac{ -106 }{ -60 }$

Simplify

$x = \dfrac{ 53 }{ 30 }$

Now that we know the value of x, we can solve for y in any of the equations, but let's use the equation "y = −5x + 8"

$\mathrm{So\:it\:would\:look\:like\:this:\ y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}$

$\mathrm{Now\:lets\:solve\:for\:"y"\:then}$

$y = -5 \left( \dfrac{ 53 }{ 30 } \right) +8}$

$\mathrm{Express\: -5 \times \dfrac{ 53 }{ 30 }\:as\:a\:single\:fraction}$

$y = \dfrac{ -5 \times 53 }{ 30 } +8$

$\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }$

$y = \dfrac{ -265 }{ 30 } +8$

$\mathrm{Simplify\: \dfrac{ -265 }{ 30 } \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }$

$y = \dfrac{ -265 \div 5 }{ 30 \div 5 } +8$

$\mathrm{Simplify}$

$y = - \dfrac{ 53 }{ 6 } +8$

$\mathrm{Turn\:8\:into\:a\:fraction\:that\:has\:the\:same\:denominator\:as\: - \dfrac{ 53 }{ 6 }}$

$\mathrm{Multiples\:of\:1: \:1,2,3,4,5,6}$

$\mathrm{Multiples\:of\:6: \:6,12,18,24,30,36,42,48}$

$\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48 }{ 6 }}$

$y = - \dfrac{ 53 }{ 6 } + \dfrac{ 48 }{ 6 }$

$\mathrm{Since\: - \dfrac{ 53 }{ 6 }\:have\:the\:same\:denominator\:,\:add\:them\:by\:adding\:their\:numerators}$

$y = \dfrac{ -53+48 }{ 6 }$

$\mathrm{Add\: -53 \: and\: 48\: to\: get\: -5}$

$y = - \dfrac{ 5 }{ 6 }$

$\mathrm{The\:solution\:is\:the\:ordered\:pair\:(\dfrac{ 53 }{ 30 }, - \dfrac{ 5 }{ 6 })}$

So there is only one solution to the equation.

5 0

2 years ago

Choosing an odd number. 1-12. Put it in a fraction

xxMikexx [17]

2/6 is a fraction. the denominator has to be an even number.

7 0

3 years ago

What are solutions for the equation y=8x-5

Semmy [17]

Answer:

Multiple

Step-by-step explanation:

You put the equation in slope intercept form, therefore, the solutions are the entire line!

Every point on the line can be considered a solution.

4 0

3 years ago

Josh is trying to factor the expression

mrs_skeptik [129]

Answer:

a) Josh likely made the negative and positive multiplication error.

b) -20a-8+12b

= -4(5a+2-3b)

Step-by-step explanation:

To check if you are correct, you can always simplify it again:

-4(5a+2-3b)

=-20a-8+12b

Remember:

Negative times negative equals to positive, negative times positive equals to negative!

I hope this is helpful! :)

4 0

3 years ago

How how many students hade more than 6 candy bars (use the picture above)​

How how many students hade more than 6 candy bars (use the picture above)