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morpeh [17]
3 years ago
12

A torch and a battery cost £2.50 altogether. The torch costs £2.00 more than the battery.

Mathematics
1 answer:
Genrish500 [490]3 years ago
8 0

Answer:

\frac{1}{10}

Step-by-step explanation:

Let t represent the cost of the torch and b represent the cost of the battery.

The torch and a battery cost £2.50 altogether.

\Rightarrow t+b=2.50

The torch costs £2.00 more than the battery.

\Rightarrow t=b+2.00

We substitute the second equation into the first equation to get;

\Rightarrow b+2.00+b=2.50

\Rightarrow 2b=2.50-2.00

£2.50

\Rightarrow 2b=0.50

\Rightarrow b=0.25

The price of the battery is £0.25

We express this as a fraction of the total cost which is £2.50 to get;

\frac{0.25}{2.5}=\frac{1}{10}

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Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav
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Answer:

  • There is no significant evidence that p1 is different than p2 at 0.01 significance level.
  • 99% confidence interval for p1-p2 is  -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

H_{0}: p1-p2=0

H_{a}: p1-p2≠0

Test statistic can be found using the equation:

z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

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  • p2 is the sample proportion of the common attribute in population2 (\frac{1337}{1900} =0.704)
  • p is the pool proportion of p1 and p2 (\frac{16+1337}{30+1900}=0.701)
  • n1 is the sample size of the people from population1 (30)
  • n2 is the sample size of the people from population2 (1900)

Then z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}} ≈ 2.03

p-value of the test statistic is  0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2±z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}} where

  • z is the z-statistic for the 99% confidence (2.58)

Thus 99% confidence interval is

0.533-0.704±2.58*\sqrt{\frac{0.533*0.467}{30}+\frac{0.704*0.296}{1900}} ≈ -0.171 ±0.237 that is (−0.408, 0.066)

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This month, Chun’s office produced 690 pounds of garbage. Chun wants to reduce the weight of garbage produced to 85% of the weig
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Chun wants to <span>reduce the weight of garbage produced to 85% of the weight produced this month. 
This means that, Chun's garbage weight next month should be equal to 85% of his garbage weight this month.

Translating this into an equation, we will find that:
garbage weight next month = 85% of garbage weight this month
garbage weight next month = 0.85 * garbage weight this month
garbage weight next month = 0.85*690 
garbage weight next month = 586.5 pounds

Based on the above calculations, his target weight for the garbage next month is </span>586.5 pounds
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Add the monthly fee $15 and the amount of times he called.

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