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katen-ka-za [31]
3 years ago
8

If the critical z-value for a hypothesis test... If the critical z-value for a hypothesis test equals 2.45, what value of the te

st statistic would provide the least chance of making a Type I error?
Mathematics
1 answer:
schepotkina [342]3 years ago
8 0

Answer: z score = 0.00714

Step-by-step explanation: the value of test statistics is gotten using the standard normal distribution table.

Z= 2.45 has area to the left (z<2.45) and area to the right (z>2.45).

Level of significance α is the probability of committing a type 1 error. The area under the distribution is known as the rejection region and it is the area towards the right of the distribution.

The table I'm using is towards the left of the distribution.

But z>2.45 + z<2.45 = 1

z> 2.45 = 1 - z<2.45

But z < 2.45 = 0.99286

z > 2.45 = 1 - 0.99286

z >2.45 = 0.00714

Hence the test statistics that would produce the least type 1 error is 0.00714

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15 hour(s) and 30 minute(s).

Step-by-step explanation:

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Its B (-1024)

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2 years ago
Help i dont understand how to complete this
Luda [366]

Answer: Josh = 19.8 hours, Danny = 10.8 hours

<u>Step-by-step explanation:</u>

Josh: \dfrac{1}{x+9}\\\\\\Danny: \dfrac{1}{x}\\\\\\Together: \dfrac{1}{7}\\\\\\Josh\quad + \quad Danny\quad =\quad Together\\\dfrac{1}{x+9}\quad +\qquad \dfrac{1}{x}\qquad = \qquad \dfrac{1}{7}\\\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (x+9)(x)(7)}}\\\\7(x) + 7(x+9)=x(x+9)\\7x + 7x + 63 = x^2+9x\\14x+63=x^2+9x\\0=x^2-5x-63\\\\\\\underline{\text{Use the quadratic formula to solve for x: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4(1)(-63)}}{2(1)}\\\\\\.\quad =\dfrac{5\pm\sqrt{25+252}}{2}\\\\\\.\quad =\dfrac{5\pm\sqrt{277}}{2}\\\\\\.\quad =\dfrac{5\pm 16.6}{2}\\\\\\x =\dfrac{5+16.6}{2}\qquad x=\dfrac{5-16.6}{2}\\\\\\x=\dfrac{21.6}{2}\qquad \qquad x=\dfrac{-11.6}{2}\\\\\\x=10.8 \qquad \qquad x=-5.8

Since time cannot be negative, x=-5.8 is an extraneous solution (not valid) so x = 10.8

Josh: x + 9   -->   10.8 + 9   =   19.8

Danny: x     -->    10.8

****************************************************************************************

2a) k = 9.45 & 0.55

2b) x = 3/2

2c) x = 2/3     <em>(x = -1 is an extraneous solution so is not valid)</em>

2d) No Solution   <em>(x = 1 is an extraneous solution so is not valid)</em>

Here is the work for 2a.  Follow this format for b, c, & d

\dfrac{3}{k^2-8x+12}=\dfrac{k}{k-2}-\dfrac{4}{k-6}\\\\\text{Since the denominator cannot equal zero, then } k \neq2\ and\ k\neq6\\\text{If either of the solutions are 2 or 6, then that solution needs to be crossed out}\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (k-2)(k-6)}}\\\\3 = k(k-6)-4(k-2)\\3=k^2-6k-4k+8\\0=k^2-10k+5\\\\\\\underline{\text{Use the quadratic formula to solve for k}}

x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4(1)(5)}}{2(1)}\\\\\\.\quad =\dfrac{10\pm\sqrt{100-20}}{2}\\\\\\.\quad =\dfrac{10\pm\sqrt{80}}{2}\\\\\\.\quad =\dfrac{10\pm8.9}{2}\\\\\\x=\dfrac{10+8.9}{2}\qquad x=\dfrac{10-8.9}{2}\\\\\\x=\dfrac{18.9}{2}\qquad x=\dfrac{1.1}{2}\\\\\\x=9.45\qquad x=0.55\\\\\\\text{Both answers are valid!}

8 0
3 years ago
HELP!!!!!:
Hatshy [7]

There are a total of 10 + 4 = 14 kg of syrup.

4 out of the 14 kg are sugar so the percentage is:

4/14 = 28.6%

Edit: The answer is correct. I don’t know why it was flagged as incorrect.

The total amount of ingredients in the syrup is 14 kg. Of the 14 kg, 4 kg is composed of sugar. So, 4 kg of sugar divided by the 14 kg of total weight will give you the percentage of sugar.

4/14 = 0.285714285714286

With an additional significant figure, the answer is 28.57%

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