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3 years ago

Will someone help me get this answer? And please explain it.

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

8 0

Do the unit rate and simplify it and. or make an equation

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3 doughnuts and 2 cakes costs £7.

Rufina [12.5K]

Answer:

cost of the doughnut = 1

cost of the cake =2

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3 years ago

Freya, a post office clerk, uses 2 yards, 1 foot, 7 inches of twine to wrap each damaged carton she repairs. She has to repair 1

Reil [10]

Answer:

42 yards 2 feet 11 inches

Step-by-step explanation:

Freya, uses 2 yards, 1 foot, 7 inches of twine to wrap each damaged carton.

She has to repair 17 cartons.

The amount of twine needed to wrap 17 cartons

= (2 yards 1 foot 7 inches) x 17

= 34 yards 17 feet 119 inches

Now let's convert the answer to larger units

1 yard = 36 inches

1 yard = 3 feet

Convert 119 inches to yards

119 inches = 3 yards 11 inches

17 feet = 5 yards 2 feet

Therefore, 34 yards 17 feet 119 inches = (34 + 3 + 5 ) yards 2 feet 11 inches

Answer = 42 yards 2 feet 11 inches

Thank you.

4 0

3 years ago

What is 1788.10643 rounded to the nearest hundredth??

wariber [46]

1788.11. Because the 0 will go to 1 because of the 6

3 0

3 years ago

Which of the following is thw missing reason?

dimulka [17.4K]

Answer:

Step-by-step explanation:

option 4

Notice that 3 is multiplied on both sides on of the equation to get

5x+6 = -9

3 0

3 years ago

Read 2 more answers

A batch of 580 containers for frozen orange juice contains 8 that are defective. Two are selected, at random, without replacemen

serious [3.7K]

Answer:

a. 0.12109

b. 0.0001668

c .0.9726

d. 0.01038

e. 0.01211

f. 0.000001731

Step-by-step explanation:

Sample size = 580

Defective units = 8

Number of picks = 2

a) If the first container is defective, there 7 defective containers left in a population of 579. The probability of selecting a defective one is:

$P=\frac{7}{579}=0.121$

b) The probability that both are defective is given by:

$P=\frac{8}{580}*\frac{7}{579}= 0.000167$

c) The probability that both are acceptable is given by:

$P=\frac{580-8}{580}*\frac{579-8}{579}= 0.9726$

d) In this case, two defective units were removed from the batch, the probability that the third is also defective is:

$P=\frac{6}{578}}= 0.0104$

e) In this case, one acceptable and one defective unit were removed from the batch, the probability that the third is also defective is:

$P=\frac{7}{578}= 0.01211$

f) The probability that all three are defective is given by:

$P=\frac{8}{580}*\frac{7}{579}*\frac{6}{578} = 0.000001731$

7 0

3 years ago