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Allushta [10]
3 years ago
5

When you roast the coffee, it loses 12% of its weight. How much fresh coffee do you need in order to get 14 225 lb. of roasted c

offee?
Mathematics
2 answers:
pochemuha3 years ago
3 0

Let f be the weight of the fresh coffee, and r be the weight of the roasted coffee. The equation described by the first sentence is

r = f\cdot \cfrac{88}{100} = f\cdot \cfrac{22}{25}

In this case, we know r and want to solve for f: the equation becomes

14225 = f\cdot \cfrac{22}{25}

To solve for f, multiply both sides by 25/22:

14225 \cdot \cfrac{25}{22}= f\cdot \cfrac{22}{25} \cdot \cfrac{25}{22}

f = 14225 \cdot \cfrac{25}{22} = \cfrac{355625}{22} \approx 16164.77

Rom4ik [11]3 years ago
3 0

Alright, lets get started.

Let me first share my double, its 14.225 lb or 14225 lb written in question.

I am considering 14.225 lb.

Lets suppose we have x lb coffee before roasting.

As given in question, it loses 12 % of its weight while roasting.

Means the rest weight of coffee after roasting = x - x * 12%

the rest weight of coffee after roasting = x - 0.12 x

the rest weight of coffee after roasting = 0.88 x

As per given in question, the weight after roasting = 14.225 lb

Equalling both

0.88x = 14.225

Dividing by 0.88

0.88x / 0.88 = 14.225 / 0.88

x = 16.1647 lb

Means we need to get 16.1647 lb coffee in order to get 14.225 lb : Answer

Hope it will help.

If 14225 lb is written in question , then calculation will be

x = 14225/0.88 = 16164.77 lb

Hope it will help :)

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Find the greatest possible error for each measurement.<br> 10 1/8 oz
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<u>Answer:</u>

0.0005 oz

<u>Step-by-step explanation:</u>

Usually, the greatest number that is allowed for approximation, assuming that the number itself is obtained by approximation, is the greatest possible error of it.

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Read 2 more answers
A bag contains 4 white,5 blacks and 2 blue balls. 3 balls are drawn one after the other without replacement from the bag .what i
Doss [256]

Answer:

24/99

Step-by-step explanation:

From the question given above, the following data were obtained:

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

Probability that they are of different color =?

Next, we shall determine the total number of balls in the bag. This can be obtained as follow:

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

TOTAL = 4 + 5 + 2 = 11 balls

Next, we shall determine the possible outcome of draw. This can be obtained as follow.

The possible outcome could be:

WBLB or WBBL or BBLW or BWBL or BLBW or BLWB

Next we shall determine the probability of each of outcomes.

Since the ball is drawn without replacement, it means the total number of ball will reduce after each draw.

White (W) balls = 4

Black (B) balls = 5

Blue (Bl) balls = 2

TOTAL = 11

P(WBLB) = 4/11 × 2/10 × 5/9 = 40/990

P(WBLB) = 4/99

P(WBBL) = 4/11 × 5/10 × 2/9 = 40/990

P(WBBL) = 4/99

P(BBLW) = 5/11 × 2/10 × 4/9 = 40/990

P(BBLW) = 4/99

P(BWBL) = 5/11 × 4/10 × 2/9 = 40/990

P(BWBL) = 4/99

P(BLBW) = 2/11 × 5/10 × 4/9 = 40/990

P(BLBW) = 4/99

P(BLWB) = 2/11 × 4/10 × 5/9 = 40/990

P(BLWB) = 4/99

Finally, we shall determine the probability that they are of different color. This can be obtained as follow:

P(WBLB) = 4/99

P(WBBL) = 4/99

P(BBLW) = 4/99

P(BWBL) = 4/99

P(BLBW) = 4/99

P(BLWB) = 4/99

Probability that they are of different color =?

Probability that they are of different color = P(WBLB) + P(WBBL) + P(BBLW) + P(BWBL) + P(BLBW) + P(BLWB)

= 4/99 + 4/99 + 4/99 + 4/99 + 4/99 + 4/99

= (4 + 4 + 4 + 4 + 4 + 4)/99

= 24/99

Thus, the probability that they are of different color is 24/99

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