All categories

2 years ago

What logarithmic function represents the data in the table?

Mathematics

2 answers:

alina1380 [7]2 years ago

8 0

216 =6^3
1,296 = 6^4
7,776 = 6^5
x = 6^f(x)
log x = log 6^f(x) = f(x)log 6
f(x) = log x / log 6 = $log_6x$

lubasha [3.4K]2 years ago

3 0

The correct answer among the choices provided is the fourth option. The logarithmic function f(x)=log(base 6)x represents the data in the table. Log is the value that one can get to express a number to a certain power. It is solved through finding the pattern using the given values,
6^3 = 216
6^4 = 1296
6^5 = 7776

You might be interested in

Express Rs 1.50 as a percentage of 200 cents

Juli2301 [7.4K]

1 $ = 100 cents

1.5 $ = 150 cents

1.50 $ a percentage of 200 cents = (150/200)100= 75%

3 0

1 year ago

If cheese costs $5.65 per pound, what will be the total cost of 2.2 pounds of cheese? plz help meee

lutik1710 [3]

The answer is $12.43

Multiply the two numbers.

3 0

2 years ago

Simplify the function f (x) = one-half (27) Superscript StartFraction 2x Over 3 EndFraction. Then determine the key aspects of t

AleksandrR [38]

Answer:

(See explanation)

Step-by-step explanation:

The function is:

$f(x) = \frac{1}{2}\cdot 27 \cdot \frac{2\cdot x}{3}$

Its simplified form is presented hereafter:

$f(x) = 9\cdot x$

The resultant expression is a linear function, first-order polynomial, whose domain and range are the set of the real numbers.

6 0

2 years ago

Read 2 more answers

Under what circumstances will the chi-square test for goodness of fit produce a large value for chi-square? Question options: a.

Whitepunk [10]

Answer:

a.when the sample proportions are much different than the hypothesized population proportions

Step-by-step explanation:

A chi-square test for goodness of fit is used to check the sample data were distributed according to claim or not.

If the chi-square test produces a large value of chi-square statistic then there is not a good fit between sample data and the null hypothesis. So, the sample proportions are much different than the hypothesized population proportions. Hence,Option (a) is correct.

If the goodness of fit produces a large value for chi-square then the sample means must not be close to the population mean. So, option (b) is incorrect.

4 0

2 years ago

Crop researchers plant 15 plots with a new variety of corn. The yields in bushels per acre are: 138.0 139.1 113.0 132.5 140.7 10

son4ous [18]

There is a relationship between confidence interval and standard deviation:
$\theta=\overline{x} \pm \frac{z\sigma}{\sqrt{n}}$
Where $\overline{x}$ is the mean, $\sigma$ is standard deviation, and n is number of data points.
Every confidence interval has associated z value. This can be found online.
We need to find the standard deviation first:
$\sigma=\sqrt{\frac{\sum(x-\overline{x})^2}{n}$
When we do all the calculations we find that:
$\overline{x}=123.8\\ \sigma=11.84$
Now we can find confidence intervals:
$($90\%,z=1.645): \theta=123.8 \pm \frac{1.645\cdot 11.84}{\sqrt{15}}=123.8 \pm5.0\\($95\%,z=1.960): \theta=123.8 \pm \frac{1.960\cdot 11.84}{\sqrt{15}}=123.8 \pm 5.99\\ ($99\%,z=2.576): \theta=123.8 \pm \frac{2.576\cdot 11.84}{\sqrt{15}}=123.8 \pm 7.87\\$
We can see that as confidence interval increases so does the error margin. Z values accociated with each confidence intreval also get bigger as confidence interval increases.
Here is the link to the spreadsheet with standard deviation calculation:
https://docs.google.com/spreadsheets/d/1pnsJIrM_lmQKAGRJvduiHzjg9mYvLgpsCqCoGYvR5Us/edit?usp=sharing

6 0

2 years ago