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3 years ago

I am having trouble with this linear equation, how do you solve this?

2 answers:

7 0

<span>3x-10y-29=0

-11x-10y+13=0

Rearrange.

</span><span>3x-10y = 0 + 29 3x - 10y = 29 ...........(i)
-
-11x-10y = 0 -13 -11x - 10y = -13.........(ii)
_____________ </span><span>
14x + 0y = 42

We subtract. Equation (i) - (ii)

3x - (-11x) = 3x + 11x = 14x

29 - (-13) = 29 + 13 = 42, -10y - (-10y) = 0

14x = 42

x = 42/14

x = 3.

Substituting x = 3, into (i), 3x - 10y = 29, 3*3 - 10y = 29

9 - 10y = 29

-10y = 29 - 9

-10y = 20

y = 20/-10

y = -2

Therefore x = 3, and y = -2.

Hope this helps. </span>

kobusy [5.1K]3 years ago

3 0

29-13=16 10-10=0y x8 16/8=-2

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Find the mean of the data summarized in the given frequency distribution . Compare the computed mean to the actual mean of 57.6

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The mean of the frequency distribution is 52.6° F. The computed mean of 52.6° F is lesser than the actual mean of 57.6° F.

<h3>What is the mean of a distribution?</h3>

The mean of the distribution can be defined as the average value of the distribution, It can be expressed as the total sum of all the observed values divided by the frequency of the distribution.

From the parameters given:

Low-temperature Frequency

40 - 44 2

45 - 49 5

50 - 54 9

55 - 59 6

60 - 64 3

The first thing to do is to determine the class midpoint. The class midpoint is the sum of the class interval divided by 2.

The class midpoint of 40 - 44 is $\mathbf{\dfrac{40 + 44}{2} = 42}$

The class midpoint of 45 - 49 is $\mathbf{\dfrac{45 + 49}{2} = 47}$

The class midpoint of 50 - 54 is $\mathbf{\dfrac{50 + 54}{2} = 52}$

The class midpoint of 55 - 59 is $\mathbf{\dfrac{55 + 59}{2} = 57}$

The class midpoint of 60 - 64 is $\mathbf{\dfrac{60 + 64}{2} = 62}$

Now, the table can be represented as:

Low-temperature Frequency x

40 - 44 2 42

45 - 49 5 47

50 - 54 9 52

55 - 59 6 57

60 - 64 3 62

The mean can now be determined as follows:

$\mathbf{\bar x = \dfrac{\sum fx}{\sum f }}$

$\mathbf{\bar x = \dfrac{(2 \times 42)+ (5 \times 47) + ( 9\times 52) +(6\times 57) + (3 \times 62)}{2 + 5 + 9 + 6 + 3 }}$

$\mathbf{\bar x = \dfrac{1315}{25}}$

$\mathbf{\bar x = 52.6}$

Therefore, we can conclude that the mean of the frequency distribution is 52.6° F. The computed mean of 52.6° F. is lesser than the actual mean of 57.6° F

Learn more about the mean of frequency distribution here:

brainly.com/question/12269435

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