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natulia [17]
3 years ago
6

Kim, Laura and Molly share £385

Mathematics
1 answer:
Feliz [49]3 years ago
3 0
We let x = Kim's amount
            y = Laura's amount
            z = Molly's amount

We have three unknowns and it should be that we have three equations to set up to be able to solve this.

x + y + z = 385    (1)
z - x = 105           (2)
2/5 = x/z              (3)

Solving simultaneously, we can get the values as follows:

x = 70
z = 175
y = 140

Therefore, the percentage that Laura got was 36.4% of the whole money. Hope this answers the question.
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If A and B are independent events, P(A) = 0.25, and P(B) = 0.3, what is P(AB)?
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Answer:

P(A) = 0.25, P(B= 0.3

And if we want to find P(A \cap B) we can use this formula from the definition of independent events :

P(A \cap B) =P(A) *P(B) = 0.25*0.3= 0.075

And the best option would be:

P(A \cap B) =0.075

Step-by-step explanation:

For this case we have the following events A and B and we also have the probabilities for each one given:

P(A) = 0.25, P(B= 0.3

And if we want to find P(A \cap B) we can use this formula from the definition of independent events :

P(A \cap B) =P(A) *P(B) = 0.25*0.3= 0.075

And the best option would be:

P(A \cap B) =0.075

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Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon en
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Answer:

Step-by-step explanation:

The null and the alternative hypothesis is:

H_o: \mu_c \ge \mu_{us}

H_a : \mu_c < \mu_{us}

The t- student test statistics can be computed as:

t = \dfrac{x_c- x_{us}}{\sqrt{\dfrac{\sigma_c^2}{n_c} + \dfrac{\sigma_{us}^2}{n_{us}} }}

t = \dfrac{19- 21}{\sqrt{\dfrac{7^2}{100} + \dfrac{8^2}{130} }}

t = -2.017  

degree of freedom = (n₁ - 1) + (n₂ - 1)  

= (100 - 1) + (130 - 1)

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Using the data of t-value and degree of freedom;  

The P-value = -0.224

Decision rule: Do not reject the null hypothesis if the p-value is greater than ∝(0.01)

Conclusion: We reject the null hypothesis since the p-value is less than ∝.

Therefore, there is enough evidence to conclude that the mean age of entering prostitution in Canada is lower than that of the United States.

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