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Luden [163]
1 year ago
10

A car company is testing a new type of tire. They want to determine the time to 60 mph from a full stop in light rain, light sno

w and dry conditions. They test out tires on five randomly selected cars from the lot. They want to know if the stopping distance is significantly different under the three conditions. What is the factor?
Mathematics
1 answer:
Andrej [43]1 year ago
3 0

Answer:

For this case the factor would be:

Road conditions

Because we are testing a new type of tire in order to determine if the time to 60 mph from a full stop in light raing, ligth snow and dry conditions.

Step-by-step explanation:

Previous concepts

By definition a factor usually known as "the independent variable is an explanatory variable manipulated by the experimenter".

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have p groups and on each group from j=1,\dots,p we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2

And we have this property

SST=SS_{between}+SS_{within}

Solution to the problem

For this case the factor would be:

Road conditions

Because we are testing a new type of tire in order to determine if the time to 60 mph from a full stop in light raing, ligth snow and dry conditions.

Th use 5 experimental units that are selected from the an specific lot. And they want to test is the stopping distance is significantly different.

And for this case we can use a one way ANOVA to test if the means are equal  in the 3 groups.

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Answer:

3/10

Step-by-step explanation:

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1 year ago
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a barangay has 1000 individuals and its population doubles every 60 years. Give an exponential model for the barangay's populati
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Answer:

P(t) = 1000e^(0.01155)t

Step-by-step explanation:

Let the population of barangay be expressed according to the exponential formula;

P(t) = P0e^kt

P(t) is the population of the country after t years

P0 is the initial population

t is the time

If barangay has 1000 initially, this means that P0 = 1000

If the population doubles after 60years then;

at t = 60, P(t) = 2P0

Substitute into the formula

2P0 = P0e^k(60)

2 = e^60k

Apply ln to both sides

ln2 = lne^60k

ln2 = 60k

k = ln2/60

k = 0.01155

Substitute k = 0.01155 and P0 into the expression

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Hence an exponential model for barangay's population is

P(t) = 1000e^(0.01155)t

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2 years ago
g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the
blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                              P.Q.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean amount of money spent on textbooks = $121.60

            \sigma = population standard deviation = $6.36

            n = sample of students = 36

            \mu = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<em />

So, 95% confidence interval for the population mean, \mu is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                      of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

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                                      = [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

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I need to find both of the solutions to the equation 100+(n-2)^ = 149
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Answer:

Step-by-step explanation:

hello :

100+(n-2)² = 149

100-100+(n-2)² = 149-100

(n-2)² = 49  

(n-2)² - 49  =0    but 49=7²

(n-2)² - 7²  =0   use identity : a²-b²=(a-b)(a+b)

(n-2-7)(n-2+7)=0

(n-9)(n+5)=0

n-9=0 or n+5=0

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