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2 years ago

By selling a book for $16.50, a bookseller loses 12%. What is the cost price of the book?

Mathematics

1 answer:

Alika [10]2 years ago

7 0

Answer: 18. 75

Step-by-step explanation:

plz follow my table format here i will be delighted to share

cp percent fraction sp

-12 88/100 16. 50

100*16.50 /88= 18.75

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Now, solve this problem. 106% of $ _______ = $7.95

Alex787 [66]

Answer:

7.5

Step-by-step explanation:

106% of $x = $7.95

In other words,

$\frac{106}{100}$ of $x = 7.95

Multiplying both sides of the equation by $\frac{100}{106}$

$\frac{106}{100} * \frac{100}{106}$ of x = 7.95 * $[\frac{100}{106}]$

=> x = $[\frac{750}{100}]$

=> x = $[\frac{7.50}{1}]$

=> x = 7.5

Validation: 106% of 7.5 = 7.95

4 0

2 years ago

Liam says - G) is equivalent to Explain if Liam is correct by completing the

vredina [299]

Answer:

The fraction 5/-12 will be shown as a positive number divided by a negative number, which always results in a negative amount. The fractions have different value, so Liam is not correct. [I'm assuming thats your fraction because its the same I had a few days ago]

8 0

2 years ago

two companies rent moving trucks. pack and go charges $39.95 flat fee and $0.45 for every mile driven. U-drive charges $59.95 fl

Nutka1998 [239]

Answer:

$62.60

Step-by-step explanation:

C(190) = 22 + 0.1*190 = 22 + 19 = $41

C(406) = 22 + 0.1*406 = 22+40.6 = $62.60

8 0

2 years ago

Read 2 more answers

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

2 years ago

a shopkeeper had purchased 400 products for $40 each the goods did not sell due to quality issues he sold them to a scrap dealer

Ratling [72]

Answer:

6500-40=6460

Step-by-step explanation:

7 0

1 year ago