Ask question

Ask question

All categories

3 years ago

12

Does the following table represent a proportional relationship, and if so what is the proportion of the cost for a box of cereal

to the size of the box?
Cost for a box of cereal $2.20 $2.64 $3.30 $4.84
Size of box (in ounces) 10 12 15 22
yes, $0.22 per ounce
yes, $4.55 per ounce
yes, $0.44 per ounce
no it is not proportional

2 answers:

mart [117]3 years ago

8 0

Answer: yes

Step-by-step explanation: 2.20 divided by 10 is .22, so one ounce is 22 cents. 2.64 divided by 12 is .22, 3.30 divided by 15 is .22, and 4.84 divided by 10 is .22. All the costs are the same per ounce, so it is a proportional relationship of .22.

elena-s [515]3 years ago

3 0

Answer:

A

Step-by-step explanation:

I'm not entirely sure but please let me know if I'm right

You might be interested in

Find the length. Round your answer to the nearest tenth

Yuri [45]

C it’s the answer I think maybe

6 0

3 years ago

The positive integer k has exactly two positive prime factors, 3 and 7. if k has a total of 6 positive factors, including 1 and

Sav [38]

I think there are two solutions, 63 or 147:

63 = 3² x 7, factors 1 | 3 | 7 | 9 | 21 | 63<span>
147 = 3 x 7</span>², factors 1 | 3 | 7 | 21 | 49 | 147

5 0

2 years ago

Will give Brainliest if you answer all

oee [108]

What grade are you in so I know what level you are on to help

7 0

2 years ago

viktelen [127]

Answer:

altemate interior angles

7 0

3 years ago

On one day a pet store sells 2 birds 6 gerbils 3 fish and 3 hamsters whats the data and relative frequency as a percent?

bonufazy [111]

Answer:

Birds $F_b=14.3\%$

Gerbils $F_g=42.9\%$

Fish $F_f=21.4\%$

Hamsters $F_h=21.4\%$

Step-by-step explanation:

From the question we are told that:

Sales

Number of birds $n_b=2$

Number of gerbils $n_g=6$

Number of fish $n_f=3$

Number of hamsters $n_h=3$

Generally the frequency for birds $F_b$ is mathematically given b

$F_b=\frac{n_b}{n}*100$

$F_b=\frac{2}{14}*100$

$F_b=14.3\%$

Generally the frequency for Gerbils $F_g$ is mathematically given b

$F_g=\frac{n_g}{n}*100$

$F_g=\frac{6}{14}*100$

$F_g=42.9\%$

Generally the frequency for fish $F_f$ is mathematically given b

$F_f=\frac{n_f}{n}*100$

$F_f=\frac{3}{14}*100$

$F_f=21.4\%$

Generally the frequency for fish $F_h$ is mathematically given b

$F_h=\frac{n_h}{n}*100$

$F_h=\frac{3}{14}*100$

$F_h=21.4\%$

5 0

2 years ago