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3 years ago

Which statement shows the associative property of addition?

Mathematics

1 answer:

abruzzese [7]3 years ago

8 0

Answer:

$\displaystyle (18 + 6) + y = 18 + (6 + y)$

Step-by-step explanation:

This shows the correct way to use the Associative Property because there are GROUPING SYMBOLS wrapped around UNIQUE parts of each expression, and STILL gets the exact same result.

I am joyous to assist you anytime.

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There are 12 girls in the swim team. they make up 48% of he students on the team. How many boys are on the swim team?

Hunter-Best [27]

Answer:

Step-by-step explanation:

12/x = 48/100

48/100 = 0.48

12 = 0.48x

12/0.48 = 25

6 0

3 years ago

Read 2 more answers

in 2009 a total of R36 000 was invested in two accounts. One account earned 7% annual interest and the other earned 9% .The tota

Akimi4 [234]

a = amount invested at 7%

b = amount invested at 9%

we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000. We can also say that

$\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{7\% of a}}{\left( \cfrac{7}{100} \right)a}\implies 0.07a~\hfill \stackrel{\textit{9\% of b}}{\left( \cfrac{7}{100} \right)b}\implies 0.09b$

since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.

$\begin{cases} a + b = 36000\\\\ 0.07a+0.09b=2920 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{a + b = 36000\implies \underline{b = 36000-a}}~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.07a~~ + ~~0.09(\underline{36000-a})~~ = ~~2920} \\\\\\ 0.07a+3240-0.09a=2920\implies 3240-0.02a=2920\implies -0.02a=-320 \\\\\\ a=\cfrac{-320}{-0.02}\implies \boxed{a=16000}~\hfill \boxed{\stackrel{36000~~ - ~~16000}{20000=b}}$

5 0

2 years ago

Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0,

Gre4nikov [31]

Answer:

C. Team Y’s scores have a lower mean value.

Step-by-step explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}$ = $\frac{26}{8}$ = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 0 + 0.25[0 - 0] = 0

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

Now, we will calculate the mean, median, range and inter-quartile range for Team Y;

Mean of Team Y data is given by the following formula;

Mean, $\bar Y$ = $\frac{\sum Y}{n}$

= $\frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}$ = $\frac{22}{8}$ = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 1 + 0.25[2 - 1] = 1.25

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.

4 0

3 years ago

When Roberto got a puppy from the shelter it weight 11 pounds. the puppy gained 40% of its original weight in the first month th

Valentin [98]

Since 40%=40/100=4/10, we can divide 11 by 10 and multiply that by 4 to get the weight gained, which is 1.1*4=4.4. The weight gained paired with the starting weight = 10+4.4=14.4

6 0

3 years ago

i need help please! I think I know the answer but for some reason my brain's telling me that it's wrong so if someone could expl

Ivenika [448]

Considering the slopes of the segments, the correct option is:

D. No, because the triangle ABC doesn't have a pair of perpendicular sides.

<h3>When are lines parallel, perpendicular or neither?</h3>

The slope, given by change in y divided by change in x, determines if the lines are parallel, perpendicular, or neither, as follows:

If they are equal, the lines are parallel.

If their multiplication is of -1, they are perpendicular.

Otherwise, they are neither.

Here, we have to find if there are perpendicular segments, that is, if two slopes multiplied have a value of -1, then:

mAB = (-7 - 9)/(11 - 1) = -8/5.

mAC = (3 - 9)/(-9 - 1) = 3/5.

mBC = (3 - (-7))/(-9 - 11) = -1/2.

No sides are perpendicular, hence option D is correct.

More can be learned about slopes at brainly.com/question/20847660

#SPJ1

6 0

2 years ago