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3 years ago

Find the interquartile range of the following data set.

Mathematics

2 answers:

sasho [114]3 years ago

8 0

Answer:

$\huge\boxed{IQR = 21}$

Step-by-step explanation:

The data set given is:

57,63,44,29,36,62,48,50,42,34

Arrange in ascending order:

29,34,36,42,44,48,50,57,62,63

Place parenthesis around the number making two equal sets.

(29,34,36,42,44) | (48,50,57,62,63)

↑ ↑

Q1 Q3

Q1 = 36 , Q3 = 57

So, IQR = Q3-Q1

IQR = 57-36

IQR = 21

sweet [91]3 years ago

7 0

Answer:

$\huge \boxed{\sf A. \ 21}$

Step-by-step explanation:

The data set is given,

$\sf 57 \ 63 \ 44 \ 29 \ 36 \ 62 \ 48 \ 50 \ 42 \ 34$

Arrange the data set in ascending order.

$\sf 29 \ 34 \ 36 \ 42 \ 44 \ 48 \ 50 \ 57 \ 62 \ 63$

Split the data set into two equal sets.

$\sf 29 \ 34 \ 36 \ 42 \ 44 \ \ \ 48 \ 50 \ 57 \ 62 \ 63$

Find the median of the lower half and upper half.

$\sf Median \ of \ lower \ half = 36$

$\sf Median \ of \ upper \ half = 57$

Interquartile range = median of upper half - median of lower half

$\sf IQR = 57 - 36$

$\sf IQR = 21$

The interquartile range for the number of points scored at ten basketball games is 21.

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lisov135 [29]

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Now:

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So:

y = 30x + 20

Question 2: Again, look for the y-intercept, which is pretty clear. It's (0, 60).

Now find two points:

(0, 60) and (2, 40)

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3 years ago

I need help with coming up with equation

Natali5045456 [20]

Let's call a child's ticket $c$ and an adult's ticket $a$ . From this, we can say:

$c + a = 116$ ,

since 116 tickets are sold in total.

Now, we are going to need to find another equation (the problem asks us to solve a systems of equations). This time, we are not going to base the equation on ticket quantity, but rather ticket price. We know that an adult's ticket is $17,000, and a child's ticket is thus

$\$17000 \cdot (1 - \dfrac{1}{4}) = \$12750$ .

Given these values, we can say:

$17000a + 12750c = 1653250$ ,

since each adult ticket $a$ costs 17,000 and each child's ticket $c$ costs 12,750, and these costs sum to 1,653,250.

Now, we have two equations:

$17000a + 12750c = 1653250$

$c + a = 116$

Let's solve:

$17000a + 12750c = 1653250$

$a = 116 - c$

Find $a$ on its own, which will allow us to substitute it into the first equation

$17000(116 - c) + 12750a = 1653250$

Substitute in $116 - c$ for $a$

$1972000 - 17000c + 12750c = 1653250$

Apply the Distributive Property

$1972000 - 4250c = 1653250$

Combine like terms

$4250c = 318750$

Subtract 1972000 from both sides of the equation and multiply both sides by -1

$c = 75$

We have now found that 75 child's tickets were sold. Thus,

$75 + a = 116 \Rightarrow a = 116 - 75 = 41$ ,

41 adult tickets were sold as well.

In sum, 41 adult tickets were sold along with 75 child tickets.

6 0

3 years ago

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