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2 years ago

PLEASE HELP I WILL GIVE YOU BRAINLIEST

Mathematics

1 answer:

DiKsa [7]2 years ago

7 0

Constant of proportionality, k, is the ratio of the quantity of variable y to the quantity of variable x.

Using a point on the line, (4, 44),

Constant of proportionality, k = .

Constant of proportionality of the graph is 11.

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If 5 burgers and 4 orders of fries cost $30.76 and 8 burgers and 6 orders of fries cost $48.28, what is the cost of a burger and

padilas [110]

Step-by-step explanation:

5B+4F= $30.76

×3

15B+12F= $92.28

----------------------------

8B+6F= $48.28

×2

16B+12F= $96.56

----------------------------

Cost of one burger:

$97.56-$92.28= $4.28 (final ans)

Cost of one fries:

$30.76-5($4.28)= $9.36 (for 5 fries)

$9.36÷5= $1.87 (final ans)

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2 years ago

you have 5 red crayons, 3 yellow crayons, 4 blue crayons and purple crayons. what multiplication would you do to find the probab

almond37 [142]

Answer:

3/46

Step-by-step explanation:

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3 years ago

Mike's Music charges $45 per hour. Tim's Tunes

klemol [59]

Answer:

Answer 4. (5 1/3 hours)

Step-by-step explanation:

240/45=5.333333333

5 0

2 years ago

Read 2 more answers

Six cards numbered from 1 to 6 are placed in an empty bowl. First one card is drawn and then put back into the bowl; then a seco

kari74 [83]

Answer:

C. $\frac{1}{18}$

Step-by-step explanation:

Given: Six cards numbered from $1$ to $6$ are placed in an empty bowl. First one card is drawn and then put back into the bowl then a second card is drawn.

To Find: If the cards are drawn at random and if the sum of the numbers on the cards is $8$ , what is the probability that one of the two cards drawn is numbered $5$ .

Solution:

Sample space for sum of cards when two cards are drawn at random is $\{(1,1),(1,2),(1,3)......(6,6)\}$

total number of possible cases $=36$

Sample space when sum of cards is $8$ is $\{(3,5),(5,3),(6,2),(2,6),(4,4)\}$

Total number of possible cases $=5$

Sample space when one of the cards is $5$ is $\{(5,3),(3,5)\}$

Total number of possible cases $=2$

Let A be the event that sum of cards is $8$

$p(\text{A})$ $=\frac{\text{total cases when sum of cards is 8}}{\text{all possible cases}}$

$p(\text{A})=\frac{5}{36}$

Let B be the event when one of the two cards is $5$

probability than one of two cards is $5$ when sum of cards is $8$

$p(\frac{\text{B}}{\text{A}})=\frac{\text{total case when one of the number is 5}}{\text{total case when sum is 8}}$

$p(\frac{\text{B}}{\text{A}})=\frac{2}{5}$

Now,

probability that sum of cards $8$ is and one of cards is $5$

$p(\text{A and B}=p(\text{A})\times p(\frac{\text{B}}{\text{A}})$

$p(\text{A and B})=\frac{5}{36}\times\frac{2}{5}$

$p(\text{A and B})=\frac{1}{18}$

if sum of cards is $8$ then probability that one of the cards is $8$ is $\frac{1}{18}$ , option C is correct.

3 0

2 years ago

Which expression finds the measure of an angle that is coterminal with a 126° angle?

N76 [4]

The expression that gives an angle that is coterminal with 126 is 126 + 720n. Two angles are said to be coterminal if when they are drawn in a standard position, their terminal sides are on the same location. The expression will give an angle which when it is drawn the terminal sides are on the same location with the 126 angle.

3 0

3 years ago

Read 2 more answers