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3 years ago

12

Samuel took an Antigua clock that his grandfather had purchased, to be appraised the appraiser evaluated the current price of th

e clock to be $400 and stated that the value will continue to increase at 2% per year. Witch of the following graphs shows the value of the antique clock, y , in dollars , after x years ?

1 answer:

NeTakaya3 years ago

4 0

For this case we have an equation of the form:
$y = A * (b) ^ x
$
Where,
A: original price
b: growth rate
x: number of years
Substituting values we have:
$y = 400 * (1.02) ^ x
$
Answer:
the value of the antique clock, and, in dollars, after x years is:
$y = 400 * (1.02) ^ x$

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Question:

Abel planned for a garden to be 42 m long but found that it needs to be 24 m long to fit in his backyard. He found the change of scale below.

$\frac{42m}{7}/{\frac{24m}{7} = \frac{6}{4}$

Which is Abel’s error?

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Abel should have divided both numbers by 6.

Abel should have divided both numbers by 4.

Answer:

Abel should have divided both numbers by 6.

Step-by-step explanation:

Given

Measured Length of backyard = 42m

Actual Length of backyard = 24m

Measured Ratio: $\frac{42m}{7}/{\frac{24m}{7} = \frac{6}{4}$

Required

Error in calculation of measured ratio

From Abel's calculation:

It can be observed that He divided both numbers by 7

Though 42/7 = 6 but that's not so for 24/7

24/7≠4;

Instead 24/7 = 3.43 (Approximated)

To get the actual ration, he has to divide both numbers by the highest common factor

The common factors of 42 and 24 include 2,3,6 of which the highest is 6

Hence; HCF of 42 and 24 = 6

So, we can conclude that the error in his calculation is having not dividing both numbers by 6

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Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

<em>Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

<u>98% confidence interval for </u> $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

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