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AleksandrR [38]
3 years ago
8

A man bought a computer on hire purchase by making a down payment of$1500 and twelve monthly payments of $250 each. calculate th

e total hire purchase price
Mathematics
1 answer:
lubasha [3.4K]3 years ago
7 0

Answer:

$4500

Step-by-step explanation:

Given parameters:

   Down payment = $1500

   Monthly payment for twelve months  =$250

Unknown;

Total hire purchase price  = ?

Solution:

To find the total hire purchase price;

  Total hire purchase price  = Down payment + 12($250)

  Total hire purchase price  = $1500 + $3000

  Total hire purchase price  = $4500

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Anika [276]

B. 3. Because if you have two people rolling a six sided die, with odd and even number contributions, you would divide 6 by 2. This is because there are TWO people going for TWO different types of numbers, odd and even. 6/2 gives us 3.


Hope this helps :)


6 0
3 years ago
Factor. 9x^2−49y^2<br><br>Enter your answer in the box
klemol [59]

Answer:

3x-7y

Step-by-step explanation:

9x^2 - 49y^2

find the square root of a values

=3x-7y

7 0
3 years ago
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ollegr [7]
The answer to part A is 1,852.50
3 0
3 years ago
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sammy [17]

Answer:

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Step-by-step explanation:

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8 0
4 years ago
The average weight of a package of rolled oats is supposed to be at most 18 ounces. A sample of 18 packages shows a mean of 18.2
Radda [10]

Answer:

Pvalue of 0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.

Step-by-step explanation:

The average weight of a package of rolled oats is supposed to be at most 18 ounces.

This means that the null hypothesis is:

H_{0}: \mu \leq 18

Is the true mean larger than the specification?

Due to the question asked, the alternate hypothesis is:

H_{a}: \mu > 18

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

18 is tested at the null hypothesis:

This means that \mu = 18

A sample of 18 packages shows a mean of 18.20 ounces with a sample standard deviation of 0.50 ounces.

This means that n = 18, X = 18.2, \sigma = 0.5

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{18.2 - 18}{\frac{0.5}{\sqrt{18}}}

z = 1.7

Pvalue of the test:

Probability of finding a sample mean above 18.2, which is 1 subtracted by the pvalue of z = 1.7.

Looking at the z-table, z = 1.7 has a pvalue of 0.9554.

1 - 0.9554 = 0.0446

0.0446 < 0.05, which means that we reject the null hypothesis and accept the alternative hypothesis, that the true mean is larger than the specification.

5 0
3 years ago
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