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1 year ago

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First 100 units = 24p per unit Remaining units = 16p per unit Mrs. Watt checks her electricity bill. Here are her meter readings

. New reading: 7143 units Old reading 6997 units (A) User subtraction to calculate the number of units used. (B) Calculate the cost of the number of units used

1 answer:

Talja [164]1 year ago

7 0

Answer: a) 146 units used b) the cost of the number of units used is 3136 p.

Step-by-step explanation:

Since we have given that

First 100 units = 24 p per unit

Remaining units = 16 p per unit

New reading = 7143 units

Old reading = 6997 units

Difference between the readings would be

$7143-6997\\\\=146$

So, there are 146 units used.

(B) Calculate the cost of the number of units used

There are 146 units

For the first 100 units = 24 p per unit

For the remaining i.e. 146-1400 = 46 units , cost = 16 p per unit

Total cost of the number of units used is given by

$24\times 100+46\times 16\\\\=2400+736\\\\=3136\ p$

Hence, a) 146 units used b) the cost of the number of units used is 3136 p.

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Answer:

At the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

Step-by-step explanation:

We are given that the mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of $2,000.

The ship building association wishes to find out whether their welders earn more or less than $20,000 annually.

<u><em>Let </em></u> $\mu$ <u><em> = mean gross annual incomes of certified welders</em></u>

So, Null Hypothesis, $H_0$ : $\mu$ = $20,000

Alternate hypothesis, $H_A$ : $\mu \neq$ $20,000

Here, null hypothesis states that the mean income of welders is equal to $20,000.

On the other hand, alternate hypothesis states that the mean income of welders is not $20,000.

Also, the test statistics that would be used here is <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean income

$\sigma$ = population standard deviation = $2,000

n = sample size

Now, at the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.

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Noor wants to spend less than $20 on condiments. Each kilogram of salt costs $1.50, and each kilogram of pepper costs 2.50.

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Step-by-step explanation:

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2 years ago

An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They

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Answer:

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Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =2.0233$ represent the sample mean 1

$\bar X_2 =3.048$ represent the sample mean 2

n1=15 represent the sample 1 size

n2=15 represent the sample 2 size

$s_1 =4.893387$ population sample deviation for sample 1

$s_2 =5.12399$ population sample deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: $\mu_1 = \mu_2$

H1: $\mu_1 \neq \mu_2$

And we can do this using the confidence interval for the difference of means.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

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$\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$

Now we have everything in order to replace into formula (1):

$-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137$

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So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

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Answer:

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Step-by-step explanation:

Given

N = ar^t

Solving (a): Write down the value of a

a implies the first term

And from the question, we understand that the initial number of trees is 5400.

Hence,

a = 5400

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Using

N = ar^t

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Substitute these values in the above expression

5184 = 5400 * r¹

5184 = 5400 * r

5184 = 5400r

Solve for r

r = 5184/5400

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First, we need to calculate number of years (t) in 2040

t = 2040 - 2014

t = 26

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