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3 years ago

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The first team member ran in a 3person relay race must run 2 1/4 laps the second member must run 1 1/2 laps the third member mus

t run 3 5/8 laps maw many laps must eatch team run

1 answer:

r-ruslan [8.4K]3 years ago

4 0

Answer:

So the total number of laps run = $7\frac{3}{8}$

Step-by-step explanation:

The first team member in a 3 person team has to run $2\frac{1}{4}$ laps, the second person has to run $1\frac{1}{2}$ and the third person has to run $3\frac{5}{8}$ laps.

We have to find total number of laps run by each team.

Total number of laps = $2\frac{1}{4}+1\frac{1}{2} +3\frac{5}{8}$

= $\frac{18}{8}+\frac{12}{8}+\frac{29}{8}$

= $\frac{59}{8}$

= $7\frac{3}{8}$

So the total number of laps run = $7\frac{3}{8}$

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The ages of students enrolled in two math classes at the local community college, Class A and Class B, are listed in order below

nlexa [21]

Answer:

The true statement about Class B is that Class B has a smaller median and the same inter quartile range.

Step-by-step explanation:

We are given the ages of students enrolled in two math classes at the local community college, Class A and Class B, below;

Class A: 20, 20, 20, 21, 22, 23, 23, 25, 27, 29, 30, 31, 34, 35, 36, 39, 40

Class B: 16, 17, 18, 18, 20, 22, 22, 24, 26, 26, 28, 29, 30, 34, 37, 40, 42

1) <u>Firstly, we will calculate Median for Class A;</u>

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

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, number of observation is odd, i.e. n = 17.

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

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 27

Hence, the median of class A is 27.

2) <u>Now, we will calculate Median for Class B;</u>

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

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, number of observation is odd, i.e. n = 17.

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

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 26

Hence, the median of class B is 26.

3) Now, we will calculate the Inter quartile range for Class A;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

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

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $21+ 0.5[22- 21]$

= 21.5

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

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $34+ 0.5[35- 34]$

= 34.5

Therefore, Inter quartile range for Class A = 34.5 - 21.5 = 13.

4) Now, we will calculate the Inter quartile range for Class B;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

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

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $18+ 0.5[20- 18]$

= 19

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

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $30+ 0.5[34- 30]$

= 32

Therefore, Inter quartile range for Class B = 32 - 19 = 13.

Hence, the true statement about Class B is that Class B has a smaller median and the same inter quartile range.

4 0

4 years ago

Find the area of the figure. Show your work. (WILL GIVE BRAINLIEST TO CORRECT ANSWER)

torisob [31]

To find the area of the figure you first have to split them up. To find the area for the triangle do 30 x 34 x 16= 16320
Then you find the area of the other shape:
30 x 30 x 16 x 40 = 576000
Then you add them together to get the answer:
576000+16320=592320
So, the answer is 592,320

I hope this helps:)

4 0

4 years ago

Read 2 more answers

A metal bar weighs 9.25 ounces. 65% of the bar is gold. how many ounces of gold is in the bar?

Elza [17]

Not sure I'm right but:

If the bar weighs 9.25 ounces and 65% of the bar is gold, you'd do 65% of 9.25. So the answer would be <span>65% of 9.25= 6.0125</span>

6 0

3 years ago

connie makes $18.25 each day she babysits if she babysits monday through friday for 4 weeks how much money will connie make

IceJOKER [234]

She workss 5 days a week*4 weeks=20 days.

20 days*$18.25 per day=$365

Final answer: $365

5 0

3 years ago

Which expression below is equivalent to 6? *<br> 1 pc<br> O 1+2+1+2+1<br> 3 (3)<br> 15-4-5<br> 24/3

Alona [7]

15-4-5
Because 15-5= 10-4 = 6

7 0

3 years ago