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2 years ago

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Grayson charges $35 per hour plus a $35 administration fee for tax preparation. Ian charges $45 per hour plus a $15 administrati

on fee. If h represents the number of hours of tax preparation, for what number of hours does Grayson charge more than Ian?
h less-than 2
h greater-than 2
h less-than 5
h greater-than 5

2 answers:

NikAS [45]2 years ago

8 0

Answer:

h less-than 2

Step-by-step explanation:

frosja888 [35]2 years ago

6 0

Answer:

h less-than 2

Step-by-step explanation:

Let

h ---> the number of hours

y ---> the total charge in dollars

we know that

The linear equation is slope intercept form is equal to

$y=mh+b$

where

m is the slope or unit rate of the linear equation

b is the y-intercept or initial value of the linear equation

we have

<em>Grayson</em>

The slope is equal to $m=\$35\ per\ hour$

The y-intercept is $b=\$35$

so

The total charge is equal to

$y=35h+35$ ----> equation A

<em>Ian</em>

The slope is equal to $m=\$45\ per\ hour$

The y-intercept is $b=\$15$

so

The total charge is equal to

$y=45h+15$ ----> equation B

For what number of hours does Grayson charge more than Ian?

The inequality that represent this situation is

$35h+35 > 45h+15$

solve for x

Group terms

$35-15 > 45h-35h$

$20 > 10h$

Divide by 10 both sides

$2 > h$

Rewrite the expression

$h< 2\ hours$ ----> h less than 2 hours

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The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.

dybincka [34]

Answer:

Critical values: $t_{\alpha/2}=-2.201$ $t_{1-\alpha/2}=2.201$

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is $\bar X= 4.059$

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n-1=12-1=11$

We can find the critical values in excel using the following formulas:

"=T.INV(0.025,11)" for $t_{\alpha/2}=-2.201$

"=T.INV(1-0.025,11)" for $t_{1-\alpha/2}=2.201$

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

And we can use Excel to calculate the limits for the interval

Lower interval : "=4.059 -2.201*(0.6503/SQRT(12))" =3.646

Upper interval : "=4.059 +2.201*(0.6503/SQRT(12))" =4.472

So the 95% confidence interval would be given by (3.646;4.472)

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One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla

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Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

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