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3 years ago

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How many extraneous solutions exist for the logarithmic equation below if it is solved in the most efficient way possible?log_(2

)[log_(2)(\sqrt(4x))]=1?

2 answers:

Nadusha1986 [10]3 years ago

7 0

Answer:

No extraneous solution

Step-by-step explanation:

We have the logarithmic equation given by,

$\log_{2}[\log_{2}(\sqrt{4x})]=1$

i.e. $\log_{2}(\sqrt{4x})=2^{1}$

i.e. $\sqrt{4x}=2^{2}$

i.e. $\sqrt{4x}=4$

i.e. $4x=4^{2}$

i.e. $4x=16$

i.e. $x=4$

So, the solution of the given equation is x=4.

Now, as we domain of square root function is x > 0 and also, the domain of logarithmic function is $( 0,\infty )$ .

Therefore, the domain of the given function is x > 0.

We know that the extraneous solution is the solution which does not belong to the domain.

But as x=4 belongs to the domain x > 0.

Thus, x = 4 is not an extraneous solution.

Hence, this equation does not have any extraneous solution.

VikaD [51]3 years ago

3 0

The correct answer on edgen is

A) 0

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Find the sum 5/14 + 4/5

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81/70

Step-by-step explanation:

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4 years ago

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In a company, 85% of the workers are women. If 390 people work for the company who aren't women, how many workers are there

aleksklad [387]

Answer:

2210

Step-by-step explanation:

390 =15%

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3 years ago

On January 1, 2017, Shay issues $700,000 of 10%, 15-year bonds at a price of 97¾. Six years later, on January 1, 2023, Shay reti

White raven [17]

Answer:

Loss on retirement is $8190

Step-by-step explanation:

In order to determine the gain or loss on the retirement of the 20% of the bonds,one needs to know the book value of the bonds retired.

First,we calculate the book value of the entire bond as follows

Initial carrying value=$700,000*97.75%=$ 684,250.00

Initial discount on bonds issue=face value -issue price

=$700,000-$684,250

=$15750

Discount amortized over 6 years out of 15 years=15750 *6/15=$6300

Unamortized discount==$15,750 -$6300=$9450

The book value of the bond now=face value-unamortized discount

=$700,000-$9450 =$690,550

book value of 20% bonds=$690550 *20%=$138,110

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3 years ago

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dolphi86 [110]

Answer:

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4 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Contact [7]

Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the 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 = $390

s = sample standard deviation = $120

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ; </u>

P(-2.101 < $t_1_8$ < 2.101) = 0.95 {As the critical value of t at 18 degrees of

freedom are -2.101 & 2.101 with P = 2.5%}

P(-2.101 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.101) = 0.95

P( $-2.101 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u> 95% confidence interval for</u> $\mu$ = [ $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }$ , $\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }$ ]

= [$332.16, $447.84]

(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion students who purchase their used textbooks = $\frac{210}{500}$ = 0.42

n = sample of students = 500

p = population proportion

<em>Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions</em>

<u>So, 99% confidence interval for the population proportion, p is ; </u>

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u> 99% confidence interval for</u> p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ , $0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ ]

= [0.363, 0.477]

Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

8 0

3 years ago