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uysha [10]
3 years ago
8

Give your answer to four decimal places

Mathematics
1 answer:
Maslowich3 years ago
4 0

Answer:

The characteristic of a logarithm is the number to the left of the decimal.

(The characteristic is like an exponent).

We are working with base 3 logs.  

So if we are increasing a number by multiplying it by the base of the logarithm, (in this case 3 times 3)  then we increase the characteristic by two.

Since 2 is the exponent of 3^2 then to get the log 3 of 72, we get the log3 (8) and increase it by 2.

1.8928 +2 = 3.8928

Had we been working with base 10 logs and we were multiplying a number by 100, We would increase the characteristic by 2 because 100 = 10^2.

Step-by-step explanation:

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PLEASE ANSWER + BRAINLIEST!!<br><br> Factor completely.<br><br> 4k - 20k^9 =<br><br> 3b^2 - 108 =
Neporo4naja [7]
4k-20k^9=4k(1-5k^8)=4k\left[1^2-(k^4\sqrt5)^2\right]\\\\=4k(1-k^4\sqrt5)(1+k^4\sqrt5)=4k\left[1^2-(k^2\sqrt[4]5)^2\right](1+k^4\sqrt5)\\\\=4k(1-k^2\sqrt[4]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k\left[1^2-(k\sqrt[8]5)^2\right](1+k^2\sqrt[4]5)(1+k^4\sqrt5)\\\\=4k(1-k\sqrt[8]5)(1+k\sqrt[8]5)(1+k^2\sqrt[4]5)(1+k^4\sqrt5)

3b^2-108=3(b^2-36)=3(b^2-6^2)=3(b-6)(b+6)

Used:\ (a-b)(a+b)=a^2-b^2
7 0
3 years ago
From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s
Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of \frac{1+\alpha}{2}.

60 out of 100 scores are passing scores, hence n = 100, \pi = \frac{60}{100} = 0.6

95% confidence level

So \alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

  • The lower limit is 0.5.
  • The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

5 0
2 years ago
15 points, please help me out
Harrizon [31]

Answer 3,5,7

Step-by-step explanation:

8 0
3 years ago
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