What is the best description of the relation in Item 2?

A 15 oz can of peaches costs $1.99 and a 24 oz can of peaches cost $2.55. How much is the price per ounce of the 15 oz can of pe

Natalka [10]

Divide the price by the number of ounces.

=$1.99 ÷ 15 ounces
=$0.13

The price per ounce of the 15 ounce can is $0.13.

Hope this helps! :)

6 0

3 years ago

Read 2 more answers

By rounding to 1 significant figure , estimate <br><br> 40.73 x 1.06<br><br> Anyone know?

suter [353]

ANSWER: 40

EXPLANATION:

1) 40.73 to 1 significant figure = 40.

2) 1.06 to 1 significant figure = 1.

3) Therefore, we just multiply 40 by 1, which equals 40.

FURTHER EXPLANATION (IF NEEDED):

With significant figures, whole numbers such as 10, 20, 50 etc. have 1 significant figure. This is because the 0 does not count as a significant figure.

With decimals, all numbers after the decimal point are also included as significant figures (even 0). E.g. 40.73 has 4 significant figures and 1.06 has 3 significant figures.

EXAMPLES (IF NEEDED):

Rounding to 2 significant figures:

40.73 = 42
1.06 = 1.1

Rounding to 3 significant figures:

40.73 = 40.7
1.06 = 1.06 (because it's already in 3 significant figures)

Rounding to 4 significant figures:

40.73 = 40.73 (already in 4 significant figures)
1.06 = 1.060 (0 is counted as a significant figure if there is a decimal or another number comes after that 0)

E.g. 1006 has 4 sig figs, while 1060 would have 3 sig figs because the last 0 is not counted as a significant figure. 10.60 would have 4 sig figs because of the decimal.

8 0

3 years ago

Dons signature coffee blend is 60% dark roast and 40% light roast. He has 10 kg of blend A, which is 80% dark roast and 20% ligh

astraxan [27]

Answer

Find out the how many kilograms of blend A will don need to use to make 10 kg of his signature blend .

To proof

As given

Dons signature coffee blend is 60% dark roast and 40% light roast.

He has 10 kg

60 % is written in the decimal form

$= \frac{60}{100}$

= 0.6

40 % is written in the decimal form

$= \frac{40}{100}$

= 0.4

Now 60% dark roast of 10 kg = 0.6 ×10

= 6 kg

Now 40% of 10 kg = 0.4 × 10

= 4 kg

As given

He has 10 kg of blend A, which is 80% dark roast and 20% light roast i.e 80% of the 6kg is dark roast and 20% of the 4kg is light roast .

80% is written in the decimal form

$= \frac{80}{100}$

= 0.8

80% of the 6kg is dark roast = 0.8 × 6

= 4.8 kg

20% is written in the decimal form

$= \frac{20}{100}$

= 0.2

20% of the 4kg is light roast = 0.2 × 4

= .8kg

Total kilograms of blend A will don need to use to make 10 kg of his signature blend = 4.8 kg + .8 kg

= 5.6 kg

Hence proved

8 0

3 years ago

Can anyone show me this in verbal form?

Andrei [34K]

Answer:

2 * (x + 2) = 50

Step-by-step explanation:

Let's call the unknown number x. "A number and 2" means that we need to add the numbers, therefore it would be x + 2. "Twice" means 2 times a quantity so "twice a number and 2" would be 2 * (x + 2). "Is" denotes that we need to use the "=" sign and because 50 comes after "is", we know that 50 goes on the right side of the "=" so the final answer is 2 * (x + 2) = 50.

4 0

3 years ago

How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits? Assume that repetitio

timofeeve [1]

Answer:

There are 6,151,600,000 different 7-place license plates are possible when 3 of the entries are letters and 4 are digits,

Step-by-step explanation:

For each of the entries which are letters, there are 26 possible outcomes.

For each of the entries which are digits, there are 10 possible outcomes.

These outcomes can be permutated.

For example, ABC1234 is a different outcome than A1B2C34. This means that we need to use the permutations formula.

The number of permutations of n, divided into two groups of size a and b, is:

$P_{a,b}^{n} = \frac{n!}{a!b!}$ .

In this problem, we have a permutation of 7, divided into a group of 4(digits) and 3(letters).

How many different 7-place license plates are possible when 3 of the entries are letters and 4 are digits?

This is $P_{3,4}^{7}*(26)^{3}*10^{4} = 35*(26)^{3}*10^{4} = 6,151,600,000$

There are 6,151,600,000 different 7-place license plates are possible when 3 of the entries are letters and 4 are digits,

7 0

3 years ago