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11111nata11111 [884]

3 years ago

11

£50 is divided between Krutika, Natasha & Richard so that Krutika gets twice as much as Natasha, and Natasha gets three time

s as much as Richard. How much does Krutika get?

2 answers:

g100num [7]3 years ago

7 0

Answer:

£30

Step-by-step explanation:

→ Model the situation in terms of x

Richard = x, Natasha = 3x and Krutika = 6x

→ Now put the values into ratio form

6x : 3x : x = 50

→ Collect the x terms

10x = 50

→ Divide both sides by 10 to isolate

x = 5

→ Substitute x = 5 into Krutika

6 × 5 = 30

AleksAgata [21]3 years ago

5 0

Answer:

£30

Step-by-step explanation:

Let's break down the information given to help us find how much Krutika gets.

<u>"Natasha gets three times as much as Richard"</u> - We can express the value that Natasha gets as $3r$ in terms of what Richard gets.

<u>"Krutika gets twice as much as Natasha"</u> - We can express this as $2(3r)$ which equals $6r$ .

We know that the total money is £50, so we can write the equation:

$r+3r+6r=50$

$10r = 50$

$r=5$

∴ Richard gets £5.

Since the money Krutika gets can be expressed as $6r$ :

$6(5) = 30$

∴ Krutika gets £30.

Hope this helps :)

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3) $\chi^2 = \frac{(21-33.143)^2}{31.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

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"=1-CHISQ.DIST(19.72,2,TRUE)"

Since the p values is lower than the significance level we have enough evidence to reject the null hypothesis at 5% of significance, and we can say that we have associated between the two variables analyzed.

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

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We need to conduct a chi square test in order to check the following hypothesis:

H0: There is significant evidence of independence between marital status and diagnostic

H1: There is significant evidence of an association between marital status and diagnostic

The level os significance assumed for this case is $\alpha=0.05$

Part 2

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

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$E_{3} =\frac{100*116}{280}=41.429$

$E_{4} =\frac{80*164}{280}=46.857$

$E_{5} =\frac{100*164}{280}=58.571$

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And the expected values are given by:

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Not Married 46.857 58.571 58.571 164

Total 80 100 100 280

Part 3

And now we can calculate the statistic:

$\chi^2 = \frac{(21-33.143)^2}{31.143}+\frac{(37-41.429)^2}{41.429}+\frac{(58-41.429)^2}{41.429}+\frac{(59-46.857)^2}{46.857}+\frac{(63-58.571)^2}{58.571}+\frac{(42-58.571)^2}{58.571} =19.72$

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Now we can calculate the degrees of freedom for the statistic given by:

$df=(rows-1)(cols-1)=(2-1)(3-1)=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >19.72)=0.00005222$

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"=1-CHISQ.DIST(19.72,2,TRUE)"

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<em></em>

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