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Kaylis [27]
3 years ago
14

Charlie bought 2 1/2 pound bag of oranges for $3.75. What is the cost of one pound of oranges

Mathematics
2 answers:
Mazyrski [523]3 years ago
4 0
Hi. The cost of a one pound bag is $1.50

To figure this out, you simply divide 2.5 into $3.75

$3.75 ÷ 2.5 lbs = $1.50

You can check your answer by adding the figures:

$1.50 for 1 lb
$1.50 for 1 lb
$0.75 (1/2 of $1.50)
$3.75 Total

Hope this helps.

Take care,
Diana
Semmy [17]3 years ago
3 0
This is how you find the answer:
Divide 3.75 /2 1/2=1.5
3.75/2.5=1.5
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Translate the sentence into an equation.
Dmitriy789 [7]

Answer:

I think it could be 8+6+y

Step-by-step explanation: I dont really know but dont delete this. Can I be Brainlyest please?

5 0
2 years ago
Please explain how to do this, I'm using all my points<br><br>A. 40 L<br>B. 80 L<br>C. 4 L <br>D. 8L
exis [7]

Answer: choice A. 40 liters

====================================

Explanation:

x = number of liters of the 50% alcohol solution

If we have x liters of 50% alcohol, then we have 0.50*x liters of pure alcohol. This is added to 0.90*40 = 36 liters of pure alcohol (from the 90% solution).

So far we have 0.50*x + 36. This expression represents the total amount of pure alcohol. We want a 70% solution, so we want 70% of the total 40+x meaning 0.50*x + 36 is to be set equal to 0.70*(40+x) and we solve for x as shown below

0.50*x + 36 = 0.70*(40+x)

0.50*x + 36 = 0.70*(40)+0.70*(x)

0.50*x + 36 = 28+0.70*x

36 - 28 = 0.70*x - 0.50x

8 = 0.20x

0.20x = 8

x = 8/0.20

x = 40

So that is why the answer is choice A. 40 liters

6 0
3 years ago
Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance
lana66690 [7]

Answer:

Null Hypothesis, H_0 : p = 0.20  

Alternate Hypothesis, H_a : p > 0.20  

Step-by-step explanation:

We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.

We have to use a 0.05 significance level to test the claim that more than 20​% of users develop nausea.

<em>Let p = population proportion of users who develop nausea</em>

So, <u>Null Hypothesis,</u> H_0 : p = 0.20  

<u>Alternate Hypothesis</u>, H_a : p > 0.20  

Here, <u><em>null hypothesis</em></u> states that 20​% of users develop nausea.

And <u><em>alternate hypothesis</em></u> states that more than 20​% of users develop nausea.

The test statistics that would be used here is <u>One-sample z proportion</u> test statistics.

                     T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }   ~ N(0,1)

where,  \hat p = proportion of users who develop nausea in a sample of 241 subjects =  \frac{54}{241}  

             n = sample of subjects = 241

So, the above hypothesis would be appropriate to conduct the test.

6 0
3 years ago
Evaluate 5j +12 foreach listed<br> value ofn.<br> j= 4<br> j = 6
Bezzdna [24]

Answer: 32 and 42

Step-by-step explanation:

5j +12

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5(4) + 12

5 times 4 = 20

20 + 12 = 32

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6 0
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Answer:

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Step-by-step explanation:

3 0
3 years ago
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