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3 years ago

(4x2y3 + 2xy2 – 2y) – (–7x2y3 + 6xy2 – 2y) i need help solving

Mathematics

1 answer:

lubasha [3.4K]3 years ago

5 0

Answer:

$\huge \boxed{11x^2 y^3-4xy^2 }$

Step-by-step explanation:

$(4x^2y^3 + 2xy^2 - 2y) - (-7x^2 y^3 + 6xy^2 -2y)$

Distribute the negative sign to the terms in the bracket.

$(4x^2y^3 + 2xy^2 - 2y) +7x^2 y^3 - 6xy^2 +2y$

Grouping like terms.

$(4x^2y^3 +7x^2 y^3) + (2xy^2 - 6xy^2 ) +( - 2y +2y)$

Combining like terms.

$(11x^2 y^3) + (-4xy^2 ) +(0)$

$11x^2 y^3-4xy^2$

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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

AlexFokin [52]

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:

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2 years ago

Does a point have a length and a width

zaharov [31]

Answer:

The answer is no.

Step-by-step explanation:

A point is neither a line or a ray, or anything that involves a line. It's just a dot, a location.

4 0

3 years ago

Read 2 more answers

What us the length of the altitude of the equilateral triangle below

ladessa [460]

Answer:

Height ( h) = 12 units.

Step-by-step explanation:

Given : An equilateral triangle with side = 8 √3.

To find : What us the length of the altitude of the equilateral triangle .

Solution : We have given equilateral triangle with side = 8 √3.

By taking the half triangle,

By the Pythagorean theorem :

(Hypotenuse)² = ( adjacent)² + (opposite)².

Plug the values Hypotenuse = 8√3 , adjacent = 4√3 , opposite = h.

(8√3)² = ( 4√3)² + (h)².

192 = 48 + (h)².

On subtracting by 48 both sides.

192 -48 = (h)².

144 = (h)².

On taking square root .

h = 12 .

Therefore, Height ( h) = 12 units.

4 0

3 years ago

To find mileage, or how many miles a car can travel per gallon of gasoline, you can use the expression m/g, where m is the dista

White raven [17]

I think the answer is 26.5 mpg. I divided 212 by 8 and got 26.5 as my answer.

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3 years ago

How to find (c,d,e)<br>Please do a step by step working.Thank you.

Natasha2012 [34]

(a)
The inverse is when you swap the variables and solve for y.
g(t) = 2t - 1 (Note: g(t) represents y)
rewrite as: y = 2t - 1
swap the variables: t = 2y - 1
solve for y: t + 1 = 2y
   $\frac{t + 1}{2}$ = y
Answer for (a): $g^{-1}(t)$ = $\frac{t + 1}{2}$

(b)
Same steps as part (a) above:
h(t) = 4t + 3
rewrite as: y = 4t + 3
swap the variables: t = 4y + 3
solve for y: $y =\frac{t - 3}{4}$

Answer for (b): $h^{-1}(t)$ = $\frac{t - 3}{4}$

(c)
$g^{-1} ( h^{-1}(t)) = g^{-1} (\frac{t - 3}{4})$
replace all t's in the $g^{-1}(t)$ equation with $\frac{t - 3}{4}$
$g^{-1} (\frac{t - 3}{4})$ = $\frac{ \frac{t-3}{4} + 1}{2}$
= $\frac{ \frac{t-3}{4} + \frac{4}{4}}{2}$ = $\frac{ \frac{t - 3 + 4}{4}}{2}$ = $\frac{ \frac{t + 1}{4}}{2} = \frac{t + 1}{8}$
Answer for (c): $g^{-1} ( h^{-1}(t))$ = $\frac{t + 1}{8}$

(d)
h(g(t)) = h(2t - 1) = 4(2t - 1) + 3 = 8t - 4 + 3 = 8t - 1
Answer for (d): h(g(t)) = 8t - 1

(e)
h(g(t)) = 8t - 1
   y = 8 t - 1
   t = 8y - 1
t + 1 = 8y
$\frac{t + 1}{8}$ = y
Answer for (e): inverse of h(g(t)) = $\frac{t + 1}{8}$

8 0

3 years ago