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levacccp [35]
3 years ago
5

Jemima bought 150 oranges at ¢12 and sold them at 6 for 30gp find the profit or loss​

Mathematics
1 answer:
stepan [7]3 years ago
8 0

Answer:

<h2>The loss is ¢ 4.50</h2>

Step-by-step explanation:

To calculate either the profit or loss we must first find the cost price and the selling price

<u>For the cost price</u>

She bought 150 oranges at ¢12

Cost price = ¢ 12.00

<u>For the selling price</u>

She sold 6 at 30 Gp that's ¢ 0.30

If she sold 6 at ¢ 0.30

Then 150 will be 150× 0.3/ 6

Selling price = ¢ 7.50

Since the cost price is higher than the selling price she made a loss

Loss = Cost price - Selling price

Loss = ¢ 12 - ¢ 7.50

<h3>Loss = ¢ 4.50</h3>

Hope this helps you

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Answer:

$192731932

Step-by-step explanation:

3 0
3 years ago
What is the ratio a:b if 8a=3b?
Digiron [165]

The ratio a:b is 3:8

Step-by-step explanation:

In simple words, the ratio a:b means the fraction a/b.

So we will have to solve the given equation for a/b

Given

8a=3b

Dividing both sides by 8

\frac{8a}{8}=\frac{3b}{8}\\a=\frac{3b}{8}

Dividing both sides by b

\frac{a}{b}=\frac{3b}{8*b}\\\frac{a}{b} =\frac{3}{8}

The ratio a:b is 3:8

Keywords: Ratios, Fractions

Learn more about ratios at:

  • brainly.com/question/4694425
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#LearnwithBrainly

5 0
3 years ago
Walter took out a $6,000 loan for six years. He is being charged 6 percent interest, compounded annually. Calculate the total am
cluponka [151]

Answer:

  $8511.11

Step-by-step explanation:

Each year, the amount Walter owes is multiplied by 1.06, so at the end of 6 years, Walter owes 1.06^6 times the amount he borrowed.

  he will pay $6,000×1.06^6 ≈ $8511.11

_____

At the end of the first year, Walter owes the original loan amount plus 6% interest. That total is ...

  $6000 + 0.06×6000 = $6000×1.06

At the end of the following year, he owes 1.06 times that amount, or ...

  6000×1.06²

The amount owed is multiplied by 1.06 each year until Walter pays off the loan.

8 0
3 years ago
A manufacturer of new light bulbs claims the average lifetime of its long-life bulb is more than 4000 hours. To test this claim,
natka813 [3]

Answer: C. 12.5

Step-by-step explanation:

Given : A manufacturer of new light bulbs claims the average lifetime of its long-life bulb is more than 4000 hours.

Population mean :  \mu=4000

Sample size : n= 100

Sample mean : \overline[x}=4500

Standard deviation: s=400

The value of test-statistic is given by :-

z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}\\\\\Rightarrow\ z=\dfrac{4500-4000}{\dfrac{400}{\sqrt{100}}}\\\\\Rightarrow\ z= 12.5

Hence, the value of the test statistic for this problem is 12.5.

5 0
3 years ago
Michmoo Computer Company sells computers and computer parts by mail. The company assures its customers that products are mailed
Anastaziya [24]

Answer: (60.858, 69.142)

Step-by-step explanation:

The formula to find the confidence interval for mean :

\overline{x}\pm z_c\dfrac{\sigma}{\sqrt{n}} , where \overline{x} is the sample mean , \sigma is the population standard deviation , n is the sample size and z_c is the two-tailed test value for z.

Let x represents the time taken to mail products for all orders received at the office of this company.

As per given , we have

Confidence level : 95%

n= 62

sample mean : \overline{x}=65 hours

Population standard deviation : \sigma=18 hours

z-value for 93% confidence interval: z_c=1.8119  [using z-value table]

Now, 93% confidence the mean time taken to mail products for all orders received at the office of this company :-

65\pm (1.8119)\dfrac{18}{\sqrt{62}}\\\\ 65\pm4.142\\\\=(65-4.142,\ 65+4.142)\\\\= (60.858,\ 69.142)

Thus , 93% confidence the mean time taken to mail products for all orders received at the office of this company : (60.858, 69.142)

8 0
4 years ago
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