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3 years ago

You prepare 10 scoops of dog food for 5 dogs, and prepare 16 scoops of dog food for 8 dogs. What

Mathematics

1 answer:

Svetllana [295]3 years ago

3 0

Answer:

12 scoops of dog food are needed for 6 dogs,

Step-by-step explanation:

Proportions

The number of scoops of dog foods and the number of dogs are proportional variables.

We are given that 10 scoops of dog foods are used for 5 dogs. This gives a proportion of 10/5=2 scoops per dog.

We also know that 16 scoops are used for 8 dogs. The proportion is also 16/8=2 scoops per dog.

Thus, the constant of proportionality is 2.

For 6 dogs, you need to prepare 2*6 = 12 scoops of dog food.

12 scoops of dog food are needed for 6 dogs,

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DedPeter [7]

Answer:

$9 \times 9 - 3 = 81 - 3 = 78$

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2 years ago

determine the mode for the following set of test scores: 63, 70, 75, 80, 85, 85, 85, 92, 92, 95, 99. (a.92) (b.85) (c.84) (d.83)

Alenkasestr [34]

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Here it is 85

Choice B

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3 years ago

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fredd [130]

Answer:

Step-by-step explanation:

Question 1: C

1 multiplied by 5 = 5

1 plus 5 = 6

So you get: (x+5)(x+1) which means x = -5 or -1

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2 multiplied by -4 = - 8

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3 years ago

Math Help???????????

RUDIKE [14]

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Step-by-step explanation:

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3 years ago

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A hospital uses cobalt-60 in its radiotherapy treatments for cancer patients. Cobalt-60 has a half-life of 7

NeTakaya

Answer:

We have 197 g of Co-60 after 18 months.

Step-by-step explanation:

We can use the decay equation.

$M_{f}=M_{i}e^{-\lambda t}$

Where:

M(f) and M(i) are the final and initial mass respectively
λ is the decay constant (ln(2)/t(1/2))
t(1/2) is the half-life of Co
t is the time at the final amount of m

$M_{f}=228e^{-\frac{ln(2)}{7} 1.5}$

$M_{f}=197\: g$

Therefore, we have 197 g of Co-60 after 18 months.

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3 years ago