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3 years ago

Can anyone help me with this?

Mathematics

1 answer:

Akimi4 [234]3 years ago

7 0

1) 2/5
2) 3/7
3) - 7/18
4) 3/5
5) - 5/18
6) 8/21
7) 2/5
8) -5
9) 9
10) 4 1/2
11) - 77/20
12) 6 10/21

By the way, to write questions 3,5 and 11, you write on the page,
(This is for Question 11)

77
- ----
20

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3 years ago

What is the logarithmic form of the solution to 102t = 9?

Harman [31]

Equivalent equations are equations that have the same value

The equation in logarithmic form is $t = \frac{\log(9)}{2}$

<h3>How to rewrite the equation</h3>

The expression is given as:

$10^{2t} = 9$

Take the logarithm of both sides

$\log(10^{2t}) = \log(9)$

Apply the power rule of logarithm

$2t\log(10) = \log(9)$

Divide both sides by log(10)

$2t = \frac{\log(9)}{\log(10)}$

Apply change of base rule

$2t = \log_{10}(9)$

Divide both sides by 2

$t = \frac{\log_{10}(9)}{2}$

Rewrite as:

$t = \frac{\log(9)}{2}$

Hence, the equation in logarithmic form is $t = \frac{\log(9)}{2}$

Read more about logarithms at:

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2 years ago

What is the interquartile range (IQR) of this data set?

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6 is the median of the lower half and 16 is the median of the upper half data scores...

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26. Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics

hjlf

The test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

<h3>What are null hypotheses and alternative hypotheses?</h3>

In null hypotheses, there is no relationship between the two phenomenons under the assumption or it is not associated with the group. And in alternative hypotheses, there is a relationship between the two chosen unknowns.

Students who take a statistics course are given a pre-test on the concepts and skills for the first chapter of a statistics course.

Then they are given a post-test once the professor has concluded lecturing on the material.

Pre-test and post-test scores for 4 students in an elementary statistics class are given below.

Then we have

$\mu _d = \mu _{post} - \mu _{pre}$

Then the null hypotheses and alternative hypotheses will be

H₀: $\mu _d = 0$

Hₐ: $\mu _d > 0$

Then the test statistic will be

$\rm \overline{x} _d = \dfrac{\Sigma x_d}{n} = \dfrac{15+12+10+1}{4}\\\\\overline{x} _d = 9.5$

Then

$\rm S_d = 6.02$

The test statistic value is given by

$\rm t = \dfrac{\overline{x} _d }{\dfrac{S_d}{\sqrtn}} \\\\t = \dfrac{9.5}{\dfrac{6.02}{\sqrt4}}\\\\t = 3.16$

Since this is a right-tailed test, so the critical value is given by

$\rm t_{n-1}(\alpha ) = t_3 (0.05) = 2.353$

Since the test statistic value lies to the right of the critical value. So we have sufficient evidence to reject the null hypothesis.

Hence, we can conclude that $\mu _d > 0$ that is test scores have improved.

More about the null hypotheses and alternative hypotheses link is given below.

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#SPJ1

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1 year ago