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Naddik [55]
3 years ago
8

Solve this equation 4log√x - log 3x =log x^2​

Mathematics
1 answer:
ruslelena [56]3 years ago
5 0

Answer:

x =  \frac{1}{3}

Step-by-step explanation:

*Move terms to the left and set equal to zero:

4㏒(√x) - ㏒(3x) - ㏒(x²) = 0

*simplify each term:

㏒(x²) - ㏒(3x) - ㏒(x²)

㏒(x²÷x²) -㏒(3x)

㏒(x²÷x² / 3x)

*cancel common factor x²:

㏒(\frac{1}{3x})

*rewrite to solve for x :

10⁰ = \frac{1}{3x}

1 = \frac{1}{3x}

1 · x = \frac{1}{3x} · x

1x = \frac{1}{3}

*that would be our answer, however, the convention is to exclude the "1" in front of variables so we are left with:

x = \frac{1}{3}

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Marine biologists have been studying the effects of acidification of the oceans on weights of male baluga whales in the Arctic O
oksano4ka [1.4K]

Answer:

a) Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that t_{\alpha/2}=1.96

And the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}  

And replacing we got:

ME=1.96\frac{125}{\sqrt{16}}=61.25  

b) ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =45 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

n=(\frac{1.960(125)}{45})^2 =29.64 \approx 30

So the answer for this case would be n=30 rounded up to the nearest integer

b. 30

c) And for this case the conditions in order to use the confidence interval are satisfied since:

a. Yes. The distribution of sample means is normal because the data are normal.

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =125 represent the population standard deviation

n=16 represent the sample size  

Part a

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that t_{\alpha/2}=1.96

And the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}  

And replacing we got:

ME=1.96\frac{125}{\sqrt{16}}=61.25  

c. 61.25 kg

Part b

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =45 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

n=(\frac{1.960(125)}{45})^2 =29.64 \approx 30

So the answer for this case would be n=30 rounded up to the nearest integer

b. 30

Part c

And for this case the conditions in order to use the confidence interval are satisfied since:

a. Yes. The distribution of sample means is normal because the data are normal.

7 0
3 years ago
Which statements about the value of the digits in the number 6,666,666 are true?
hammer [34]
I believe c and d are correct
5 0
3 years ago
Use the rectangle to answer the question.
nikitadnepr [17]

Answer:

12 x 9 = 108

Step-by-step explanation:

Just multiply the 2 given lengths when finding AREA of a rectangle.

7 0
3 years ago
Read 2 more answers
How many tenths are there in 4/5
Minchanka [31]
8 tenth because 4 over 5 is also 8 over 10
7 0
3 years ago
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This is just a question I want u guys to answer cause 99% of the adult population don't know the answer to this question.
Marrrta [24]
50/2 = 25

also can i have a brainliest
6 0
3 years ago
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