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topjm [15]
2 years ago
11

Which statement is correct? StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFrac

tion less-than-or-equal-to (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline) StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFraction less-than (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline) StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFraction greater-than (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline) StartFraction 3.56 times 10 Superscript 2 Baseline Over 1.09 times 10 Superscript 4 Baseline EndFraction = (4.08 times 10 Superscript 2 Baseline) (1.95 times 10 Superscript negative 6 baseline)
Mathematics
2 answers:
erma4kov [3.2K]2 years ago
8 0

The answer is C.

I hope this helps :)

stellarik [79]2 years ago
5 0
Mannn what the heck...it’s so confusing sorry I couldn’t read.its messing
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A facilities manager at a university reads in a research report that the mean amount of time spent in the shower by an adult is
MakcuM [25]

Answer:

We conclude that the mean amount of time that college students spend in the shower is equal to 5 minutes.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 5 minutes

Sample mean, \bar{x} = 4.61 minutes

Sample size, n = 15

Alpha, α = 0.05  

Sample standard deviation, s = 0.75 minutes

a) First, we design the null and the alternate hypothesis  

H_{0}: \mu = 5\text{ minutes}\\H_A: \mu \neq 5\text{ minutes}  

We use two-tailed t test to perform this hypothesis.  

b) Formula:  

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have  

t_{stat} = \displaystyle\frac{4.61 - 5}{\frac{0.75}{\sqrt{15}} } = -2.0139  

c) Now, we calculate the p-value using the standard table.

P-value = 0.0638

d) Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

We conclude that the mean amount of time that college students spend in the shower is equal to 5 minutes.

Option a) Fail to reject the claim that the mean time is 5 minutes because the P-value is larger than 0.05.

4 0
1 year ago
mrs bean bought a business at the start of 2006 the business was then valued £30000 Since Mrs Beans takeover, the business gas c
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Answer:

38000(Rounded to nearest 100)

Step-by-step explanation:

2018-2006=12 30000X1.02^12 =38047.2538

4 0
2 years ago
Marquis used the steps below to find the solution to the inequality Negative 5.6 greater-than-or-equal-to x + 3.6.
Taya2010 [7]

Answer:3.9

Step-by-step explanation:

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5 0
2 years ago
Read 2 more answers
Pierce currently has $10,000. What was the value of his money five years ago if he has earned 5 percent interest each year?
tia_tia [17]
The first answer is correct, if you go back by 5% each year you will see that.

4 0
2 years ago
Read 2 more answers
The heat evolved in calories per gram of a cement mixture is approximately normally distributed. The mean is thought to be 100,
Gre4nikov [31]

Answer:

A.the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

\mathtt{H_o: \mu = 100}

\mathtt{H_1: \mu \neq 100}

A. If the acceptance region is defined as 98.5 <  \overline x >  101.5 , find the type I error probability \alpha .

Assuming the critical region lies within \overline x < 98.5 or \overline x > 101.5, for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is \mu = 100

∴

\mathtt{\alpha = P( type  \ 1  \ error ) = P( reject \  H_o)}

\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5  )}

when  \mu = 100

\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}}  > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }

\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z  > \dfrac{1.5}{\dfrac{2}{3}}) }

\mathtt{\alpha = P ( Z  2.25) }

\mathtt{\alpha = P ( Z

From the standard normal distribution tables

\mathtt{\alpha = 0.0122+( 1-  0.9878) })

\mathtt{\alpha = 0.0122+( 0.0122) })

\mathbf{\alpha = 0.0244 }

Thus, the type 1 error probability is \mathbf{\alpha = 0.0244 }

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis \mathtt{H_o}

Thus;

β = P( type II error) - P( fail to reject \mathtt{H_o} )

∴

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 103

\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }

\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

\mathtt{\beta = P(98.5 \leq \overline x \leq  101.5)           }

Given that \mu = 105

\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }

\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }

\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }

\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

8 0
2 years ago
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