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3 years ago

8

The cost C , in £, of a monthly phone contract is made up of the fixed line rental L , in £, and the price P , in £ ,of the call

s made. Enter a formula for the cost and, enter the cost if the line rental is £10 and the price of calls made is £28.

1 answer:

Mama L [17]3 years ago

3 0

Answer:

C = L + P

C = £38

Step-by-step explanation:

Given that :

Cost (C) is given by :

Sum of fixed line rental (L) ; price (P) of calls made.

Hence,

Cost formula;

C = L + P

Cost if the line rental is £10 and the price of calls made is £28.

C = L + P

If L = £10 ; P = £28

C = £10 + £28

C = £38

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Answer:

The question is incomplete, but the step-by-step procedures are given to solve the question.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.005 = 0.995$ , so Z = 2.575.

Now, find the margin of error M as such

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In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

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The lower end of the interval is the sample mean subtracted by M.

The upper end of the interval is the sample mean added to M.

The 99% confidence interval for the population mean amount of beverage in 16-ounce beverage cans is (lower end, upper end).

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Answer:

1. Significantly different from

1. Inconsistent

Step-by-step explanation:

Midwest region total stores equal to 63.1%. The percentage for current store is less than total stores therefore data in the given table is not consistent. The data given is relatively for the first month sales and is significantly different from the other stores.

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Step-by-step explanation:

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An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

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The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

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