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vaieri [72.5K]
1 year ago
13

The limits of the class intervals are

Mathematics
1 answer:
Step2247 [10]1 year ago
6 0

The lower and the upper class limit are as follows

interval  lower limit  upper limit

10-14          10                14

15-19          15                19

20-24         20              24

25-29           25            29

This is further explained below.

<h3>What is the class limit?</h3>

Generally, The lower class limit and the upper-class limits are just the minima and maximum values that are allowed in each class, respectively: The gap that exists between the upper-class limit and the lower class limit is referred to as the class interval.

In conclusion, The lower and the upper class limit are as follows

interval  lower limit  upper limit

10-14          10                14

15-19          15                19

20-24         20              24

25-29           25            29

Read more about the class limit

brainly.com/question/15259987

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Answer:

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Step-by-step explanation:

Let x represent the number of sales each man had.

For Salesman A, he earns $65 per sale; this is 65x.

For Salesman B, he earns $40 per sale; this is 40x.  We also add to this his weekly salary of $300; this gives us 40x+300.

Since their pay was equal, set the two expressions equal:

65x = 40x+300

Subtract 40x from each side:

65x-40x = 40x+300-40x

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"Select the equation of the least squares line for the data: (34.0, 1.30), (32.5, 3.25), (35.0, .65), (31.0, 6.50), (30.0, 5.85)
yulyashka [42]

Answer:

y=-1.055 x +37.643

And the best option would be:

a. ŷ= 37.643-1.0543x

Step-by-step explanation:

We assume that the data is this one:

x: 34.0, 32.5, 35.0, 31.0, 30.0, 27.5, 29.0

y: 1.30, 3.25, 0.65, 6.50, 5.85, 8.45, 6.50

Find the least-squares line appropriate for this data.  

For this case we need to calculate the slope with the following formula:

m=\frac{S_{xy}}{S_{xx}}

Where:

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}

So we can find the sums like this:

\sum_{i=1}^n x_i = 34.0+ 32.5+ 35.0+ 31.0+ 30.0+ 27.5+ 29.0&#10;=219

\sum_{i=1}^n y_i =1.3+3.25+0.65+6.5+5.85+8.45+6.5=32.5

\sum_{i=1}^n x^2_i = 34.0^2+ 32.5^2+ 35.0^2+ 31.0^2+ 30.0^2+ 27.5^2+ 29.0^2&#10;=6895.5

\sum_{i=1}^n y^2_i =1.3^2+3.25^2+0.65^2+6.5^2+5.85^2+8.45^2+6.5^2=202.8

\sum_{i=1}^n x_i y_i =34*1.3 + 32.5*3.25+ 35*0.65+ 31*6.5+30*5.585 +27.5*8.45 +29*6.5=970.45

With these we can find the sums:

S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=6895.5-\frac{219^2}{7}=43.929

S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=970.45-\frac{219*32.5}{7}=-46.336

And the slope would be:

m=-\frac{46.336}{43.929}=-1.055

Nowe we can find the means for x and y like this:

\bar x= \frac{\sum x_i}{n}=\frac{219}{7}=31.286

\bar y= \frac{\sum y_i}{n}=\frac{32.5}{7}=4.643

And we can find the intercept using this:

b=\bar y -m \bar x=4.643-(-1.055*31.286)=37.643

So the line would be given by:

y=-1.055 x +37.643

And the best option would be:

a. ŷ= 37.643-1.0543x

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Answer:

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