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Angelina_Jolie [31]
3 years ago
9

B) y = log2 (4" - 4) find the inverse of this equation

Mathematics
1 answer:
GuDViN [60]3 years ago
4 0

Answer:

y =  \frac{1}{4}  \times  {2}^{x}  - 1

Step-by-step explanation:

Assuming the given logarithmic equation is

y =   \log_{2}( {4}{x }  - 4)

We interchange x and y to get:

x=   \log_{2}( {4}{y} - 4)

We solve for y now:

{2}^{x}  =  {4}{y}   - 4

We add 4 to both sides to get;

{2}^{x}  + 4 = 4y

Divide through by 4:

y =  \frac{1}{4}  \times  {2}^{x}  - 1

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Step-by-step explanation:

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2-9%4*2^4<br><br><br> Undyne with a Mohawk<br> Not my best work but at least I did it
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Step-by-step explanation:

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3 years ago
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The national mean annual salary for a school administrator is $90,000 a year (The Cincinnati Enquirer, April 7, 2012). A school
aliina [53]

Answer:

Step-by-step explanation:

The question is incomplete. The missing part is given below:

A) Formulate hypotheses that can be used to determine whether the population mean annual administrator salary in Ohio differs from the national mean of $90,000.

B) The sample data for 25 Ohio administrators is contained below. What is the p-value for your hypothesis test in part A?

C) At alpha = 0.05, can your null hypothesis be rejected? What is your conclusion?

D) Repeat the preceding hypothesis test using the critical value approach?

Salary Data

77600 , 76000 , 90700 , 97200 , 90700 , 101800 , 78700 , 81300 , 84200 , 97600 , 77500 , 75700 , 89400 , 84300 , 78700 , 84600

, 87700 , 103400 , 83800 , 101300 , 94700 , 69200 , 95400 , 61500 , 68800

Solution:

Mean = total sum of salaries/number of school administrators

Mean = (77600 + 76000 + 90700 , 97200 + 90700 + 101800 + 78700 + 81300 + 84200 + 97600 + 77500 + 75700 + 89400 + 84300 + 78700 + 84600 + 87700 + 103400 + 83800 + 101300 + 94700 + 69200 + 95400 + 61500 + 68800)/25 = 85272

Standard deviation = √(summation(x - mean)²/n

n = 25

√Summation(x - mean)²/n = √[(77600 - 85272)² + (76000 - 85272)² + (90700 - 85272)² + (97200 - 85272)² + (90700 - 85272)² + (101800 - 85272)² + (78700 - 85272)² + (81300 - 85272)² + (84200 - 85272)² + (97600 - 85272)² + (77500 - 85272)² + (75700 - 85272)² + (89400 - 85272)² + (84300 - 85272)² + (78700 - 85272)² + (84600 - 85272)² + (87700 - 85272)² + (103400 - 85272)² + (83800 - 85272)² + (101300 - 85272)² + (94700 - 85272)² + (69200 - 85272)² + (95400 - 85272)² + (61500 - 85272)² + (68800 - 85272)²]/25 = 11039.23

A) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0 : µ = 90000

For the alternative hypothesis,

H1 : µ ≠ 90000

This is a 2 tailed test.

B) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 25

Degrees of freedom, df = n - 1 = 25 - 1 = 24

t = (x - µ)/(s/√n)

Where

x = sample mean = 85272

µ = population mean = 90000

s = samples standard deviation = 11039.23

t = (85272 - 90000)/(11039.23/√25) = - 2.14

We would determine the p value using the t test calculator. It becomes

p = 0.043

C) Since alpha, 0.05 > the p value, 0.043, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the population mean annual administrator salary in Ohio differs from the national mean of $90,000.

D) Since α = 0.05, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.05/2 = 0.025

The z score for an area to the left of 0.005 is - 1.96

For the right, α/2 = 1 - 0.025 = 0.975

The z score for an area to the right of 0.975 is 1.96

In order to reject the null hypothesis, the test statistic must be smaller than - 1.96 or greater than 1.96

Since - 2.14 < - 1.96 and 2.14 > 1.96, we would reject the null hypothesis.

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Answer:

the answer : a

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6 + 5|2x -3| ≥ 4 pls pls pls help
sattari [20]

Answer:

x is all real numbers

Step-by-step explanation:

6 + 5|2x -3| ≥ 4

The first step is to subtract 6 from each side

6 -6+ 5|2x -3| ≥ 4-6

5|2x -3| ≥ -2

Now divide each side by 5

5/5|2x -3| ≥ -2/5

|2x -3| ≥ -2/5

Absolute values are always positive.  Our absolute value is greater than a negative number.  It is always true.  X can be any number and our absolute value will still be greater than -2/5.

x is all real numbers

4 0
3 years ago
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