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lana66690 [7]
3 years ago
15

Pick 1 pls... im not trynna be mean but pls dont put an answer up here if u dont know

Mathematics
2 answers:
sveta [45]3 years ago
6 0

Answer:

It is the first one

Step-by-step explanation:

mart [117]3 years ago
3 0
I believe it is number 1
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Express f(x) = log(100) in the general form of a logarithmic function, g(x) = k + a log b(r – h).
agasfer [191]

The logarithmic equation f(x) = log(100) in the form g(x) = k + a log b(r – h) is f(x) = 0 + 2log(10 - 0)

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

The equation is given as:

f(x) = log(100)

Express 100 as 10^2

f(x) = log(10^2)

Apply the law of logarithm

f(x) = 2log(10)

Subtract 0 from 10

f(x) = 2log(10 - 0)

Add 0 to the equation

f(x) = 0 + 2log(10 - 0)

Hence, the logarithmic equation f(x) = log(100) in the form g(x) = k + a log b(r – h) is f(x) = 0 + 2log(10 - 0)

Read more about logarithmic equation at:

brainly.com/question/13473114

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6 0
2 years ago
Rewrite the expression using rational exponents.
Vedmedyk [2.9K]

Answer:

x^{8/3}

Step-by-step explanation:

when taking the radical of a number its really just a fractioned exponent for example \sqrt{x} = x^{1/2}

This is because in the radical the x has an exponent of 1 and since its the square root the radical is to the 2nd so you take the exponent of the X and put that of the what ever kind of radial you have such as a square root third root fourth root and so on...

in your case you would take X's exponent of 8 and since your taking the cube root would put the 8 over the 3.

hope that helped

4 0
3 years ago
Read 2 more answers
A researcher wants to show the mean from population 1 is less than the mean from population 2 in​ matched-pairs data. if the obs
Veronika [31]

Answer:

The null hypothesis is represented as

H₀: μd = 0

The alternative hypothesis is represented as

H₁: μd < 0, that is, mud less than 0.

Step-by-step explanation:

Correct Question

A researcher wants to show the mean from population 1 is less than the mean from population 2 in​ matched-pairs data. if the observations from sample 1 are xi and the observations from sample 2 are yi​, and di = (xi -yi​), then the null hypothesis is h0​: μd = 0 and the alternative hypothesis is h1​: μd ​___ 0.

Solution

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

In hypothesis testing, especially one comparing two sets of data, the null hypothesis plays the devil's advocate and usually takes the form of the opposite of the theory to be tested. It usually contains the signs =, ≤ and ≥ depending on the directions of the test. It usually maintains that, with random chance responsible for the outcome or results of any experimental study/hypothesis testing, its statement is true.

The alternative hypothesis usually confirms the theory being tested by the experimental setup. It usually contains the signs ≠, < and > depending on the directions of the test. It usually maintains that significant factors other than random chance, affect the outcome or results of the experimental study/hypothesis testing and result in its own statement.

For this question, we are aiming to show that the mean from population 1 (μₓ) is less than the mean from population 2 (μᵧ).

The null hypothesis would be that there isn't significant evidence to suggest that the mean from population 1 (μₓ) is less than the mean from population 2 (μᵧ). That is, the mean from population 1 (μₓ) is not less than the mean from population 2 (μᵧ).

The alternative hypothesis is that there is significant evidence to suggest that mean from population 1 (μₓ) is less than the mean from population 2 (μᵧ).

Mathematically, if μₓ and μᵧ are the means of population 1 and 2 respectively and μd is the mean of the differences between population 1 and population 2, μd = μₓ - μᵧ

The null hypothesis is represented as

H₀: μₓ = μᵧ or μd = 0

The alternative hypothesis is represented as

H₁: μₓ < μᵧ or μd < 0

Hope this Helps!!!

8 0
3 years ago
Suppose we roll a fair die and let X represent the number on the die. (a) Find the moment generating function of X. (b) Use the
Likurg_2 [28]

Answer:

(a)  moment generating function for X is \frac{1}{6}\left(e^{t}+e^{2 t}+e^{2 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)

(b) \mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}

Step-by step explanation:

Given X represents the number on die.

The possible outcomes of X are 1, 2, 3, 4, 5, 6.

For a fair die, P(X)=\frac{1}{6}

(a) Moment generating function can be written as M_{x}(t).

M_x(t)=\sum_{x=1}^{6} P(X=x)

M_{x}(t)=\frac{1}{6} e^{t}+\frac{1}{6} e^{2 t}+\frac{1}{6} e^{3 t}+\frac{1}{6} e^{4 t}+\frac{1}{6} e^{5 t}+\frac{1}{6} e^{6 t}

M_x(t)=\frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right)

(b) Now, find E(X) \text { and } E\((X^{2}) using moment generating function

M^{\prime}(t)=\frac{1}{6}\left(e^{t}+2 e^{2 t}+3 e^{3 t}+4 e^{4 t}+5 e^{5 t}+6 e^{6 t}\right)

M^{\prime}(0)=E(X)=\frac{1}{6}(1+2+3+4+5+6)  

\Rightarrow E(X)=\frac{21}{6}

M^{\prime \prime}(t)=\frac{1}{6}\left(e^{t}+4 e^{2 t}+9 e^{3 t}+16 e^{4 t}+25 e^{5 t}+36 e^{6 t}\right)

M^{\prime \prime}(0)=E(X)=\frac{1}{6}(1+4+9+16+25+36)

\Rightarrow E\left(X^{2}\right)=\frac{91}{6}  

Hence, (a) moment generating function for X is \frac{1}{6}\left(e^{t}+e^{2 t}+e^{3 t}+e^{4 t}+e^{5 t}+e^{6 t}\right).

(b) \mathrm{E}(\mathrm{X})=\frac{21}{6} \text { and } E\left(X^{2}\right)=\frac{91}{6}

6 0
4 years ago
2*2=4-1=A<br> What does a equal
zmey [24]

Answer:

Step-by-step explanation: pemdas

4 divided by 4 - 1

1-1=0

a = 0

8 0
3 years ago
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