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daser333 [38]
3 years ago
6

If a 3-pound bunch of bananas cost $1.50, how

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
4 0

Answer:

Approximately 3 bunches would equal $4.50. So you should be able to buy at least 4 bunches which would be 10 pounds

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Suppose you have $10 and your brother has $54. If you decide to save $10 of your allowance each week and your brother decides to
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Answer:  11 weeks

Step-by-step explanation: 11 x 10 + 10 = $120 and 11 x 6 +54 = $120

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What's an excuse if you miss a class​
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Answer:

Say you ate something bad and you kept throwing up idrk

Step-by-step explanation:

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) You are finally well!  But now your child is sick!  You know there have been mistakes in orders so you are on the internet t
rjkz [21]

Answer:

The doctor's dosage was not appropriate.

The right dosage is from 56.625 mg to 113.25 mg of Antibiotic to be given every 6 hour for the child with weight 25 lbs.

Step-by-step explanation:

It is given that 20 to 40 mg/kg/day is the recommended dosage.

Now, the doctor's order is 150 mg of antibiotic to be given every 6 hours. And my child weighs 25 lbs.

Now, 1 lb is equivalent to 0.453 kg.

So, my child's weight is (25 × 0.453) = 11.325 kg.

So, the doctor's order is 150 mg of antibiotic to be given every 6 hours for a child of weight 11.325 kg.

Hence, the dosage is [(150 × 4) ÷ 11.325] = 52.98 mg/kg/day.

So, this is not within the limit of 20 to 40 mg/kg/day.

Therefore, the doctor's dosage was not appropriate.

Now, let the appropriate dosage is x mg per every 6 hours for a child with weight 25 lbs i.e. 11.325 kg.

So, 20 \leq  \frac{4x}{11.325} \leq 40

⇒ 20\leq 0.3532x \leq 40

⇒ 56.625 \leq  x \leq  113.25

So, the right dosage is from 56.625 mg to 113.25 mg of antibiotic to be given every 6 hours for the child with weight 25 lbs. (Answer)

5 0
3 years ago
Which ratio cannot form a proportion with 3/4?<br><br>A.)30/40<br>B.)12/20<br>C.)12/16<br>D.)9/12
Vesna [10]
The answer would be B)
4 0
3 years ago
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A pile of coins, consisting of quarters and half dollars, is worth $11.75. If there are 2 more quarters than half dollars, how m
Black_prince [1.1K]

The pile contains 17 quarters and 15 half-dollars.

Let <em>x</em> = the number of quarters and <em>y</em> = the number of half-dollars.

We have two equations:

(1) $0.25<em>x</em> + $0.50<em>y</em>  = $11.75

(2) <em>x</em> = <em>y</em> +2

Substitute the value of <em>x</em> from Equation (2) into Equation (1).

0.25(<em>y</em>+2) + 0.50<em>y</em> = 11.75

0.25<em>y</em> + 0.50 + 0.50<em>y</em> = 11.75

0.75<em>y</em> = 11.75 – 0.50 = 11.25

<em>y</em> = 11.25/0.75 = 15

Substitute the value of <em>y</em> in Equation (2).

<em>x</em> = 15 + 2 = 17

The pile contains 17 quarters and 15 half-dollars.

<em>Check</em>: 17×$0.25 + 15×$0.50 = $4.25 + $7.50 = $11.75.

4 0
3 years ago
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