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3 years ago

Question is based on Fraction

Mathematics

1 answer:

Genrish500 [490]3 years ago

6 0

9514 1404 393

Answer:

12,00,000 rupees

Step-by-step explanation:

The fraction originally intended for each son was ...

(1/3)(1 -2/5) = 1/3(3/5) = 1/5

The fraction each son actually received was ...

(1 -1/6)(1/5) = (5/6)(1/5) = 1/6

The difference in these amounts as a fraction of the original distribution is ...

1/5 - 1/6 = (6/30) -(5/30) = 1/30

Then the original distribution is 30 times this difference:

30×40000 rupees = 12,00,000 rupees

_____

Additional comment

Originally-intended distribution (rupees):

wife: 4,80,000
each son: 2,40,000

Actual distribution:

brother: 2,00,000
wife: 4,00,000
each son: 2,00,000

1,00,000 = 1 lakh

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7xy + 8z - 2z + 12xy + 10z

cestrela7 [59]

Answer:

19xy + 16z

Step-by-step explanation:

combine like terms

7xy +1 2xy= 19xy

8z - 2z + 10z = 16z

7 0

3 years ago

Combine the like terms to create an equivalent expression: −4p+(−6p)

Y_Kistochka [10]

THE ANSWER IS -10P.

Step-by-step explanation:

USING BY MULTIPLICATION RULE.

minus × plus = minus

then minus × minus = plus.

so we add the 4p+6p= -10p.

why -10p is come it means THE BIGGER NUMBER SIGN IS MINUS. So it's comes -10p.

HOPE IT HELP......

7 0

3 years ago

In the 2009-2010 school year in country A, there were 92,000 foreign students from country B. This number I 23% more than the nu

max2010maxim [7]

Answer:

$\approx74796$

Step-by-step explanation:

GIVEN: In the $2009-10$ school year in country A, there were $92,000$ foreign students from country B. This number is $23\%$ more than the number of students from country C.

TO FIND: How many foreign students were from country C.

SOLUTION:

Total foreign students from country B $=92000$

let the foreign students from country C be $x$

As the students from country B is $23\%$ more than the country C

Students from country B

$92000=\frac{123}{100}x$

$x=\frac{9200000}{123}$

$x\approx74796$

Hence the number of students from country C are $74796$

8 0

3 years ago

An average computer mouse inspector can inspect 50 mice per hour. The 48 computer mice inspectors at a particular factory can on

Ronch [10]

Answer: C. Yes, because -2.77 falls in the critical region .

Step-by-step explanation:

Let $\mu$ be the population mean .

As per given , we have

$H_0:\mu=50\\\\ H_a: \mu$

Since the alternative hypothesis is left-tailed and population standard deviation is not given , so we need to perform a left-tailed t-test.

Test statistic : $t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

Also, it is given that ,

n= 48

$\overline{x}=46$

s= 10

$t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}=\dfrac{-4}{\dfrac{10}{6.93}}\\\\=\dfrac{-4}{1.443}\approx-2.77$

Degree of freedom = df = n-1= 47

Using t-distribution , we have

Critical value = $t_{\alpha,df}=t_{0.025,47}=2.0117$

Since, the absolute t-value (|-2.77|=2.77) is greater than the critical value.

So , we reject the null hypothesis.

i.e. -2.77 falls in the critical region.

[Critical region is the region of values that associates with the rejection of the null hypothesis at a given probability level.]

Conclusion : We have sufficient evidence to support the claim that these inspectors are slower than average.

Hence, the correct answer is C. Yes, because -2.77 falls in the critical region

4 0

3 years ago

(08.03)jasper is using the following data samples to make a claim about the house values in his neighborhood: house value a $150

ipn [44]

Part A) Jasper should use the median because there is an outlier that affects the mean is the correct answer.

Part B) Option (D) no, because there is a lot of variability in the art book data is the correct answer.

In this question,

Part A)

The median is the middle number in a sorted, ascending or descending list of numbers and can be more descriptive of that data set than the average. The numbers are ordered in ascending order. Once the numbers are ordered, the middle number is called the median of the given data set.

The median is important because it gives us an idea of where the center value is located in a dataset. The median is the most informative measure of central tendency for skewed distributions or distributions with outliers.

The house values in Jasper's neighborhood,

a) $150,000, b) $175,000, c) $200,000 ,d) $167,000, e) $2,500,000

The value of house E is very high from other houses in Jasper's neighborhood. The value of house E will make the mean house value above the center of the data as House E is an outlier for our given data.

Thus the median of a data set with an outlier is not affected by the value of outlier. Therefore, Jasper should use the median because there is an outlier that affects the mean is the correct answer.

Part B)

The mean absolute deviation of a dataset is the average distance between each data point and the mean. It gives us an idea about the variability in a dataset. The mean absolute deviation is specially important in data analysis where the average value can be deceptive.

A small mean absolute deviation tells us that most of the data values are very close to the mean. A high mean absolute deviation tells us that many of the data values are spread out further from the mean.

Magdeline selects 25 pages from each book, then Calculates Mean, Median, Mean Absolute Deviation of number of words on each page.

Book Mean Median Mean Absolute Deviation

Art 68.7 50 15.4

Science 54.2 55 7.9

Magdeline claims that the mean number of words on each page in the art book is greater than the mean number of words on each page in the science book, the art book has more words per page is not a valid inference.

Because variability from mean in art book is more than variability from mean in science book and the mean absolute deviation measures variability from mean. Thus option (D) no, because there is a lot of variability in the art book data is the correct answer.

Learn more about mean absolute deviation here

brainly.com/question/1468924

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7 0

2 years ago