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2 years ago

13

PLZ HELP ME ASAP The box plot shows the typing speed (in words per minute without errors) of the contestants in a typing contest

.
The interquartile range of team A is
A.3
B.4
C.8
D.9
And the interquartile range of team B is
A.4
B.5
C.8
D.9
The difference of the medians of team A and team B is
A.4
B.7
C.8
D.9
This value is equal to about
A.half
B.1 times
C.2 times
the interquartile range of either data set.

2 answers:

Jet001 [13]2 years ago

8 0

Answer:

1. Option C is correct.

2. Option D is correct.

3. Option A is correct.

4. Option A is correct.

Step-by-step explanation:

Starting and end point of box plot represent the minimum and maximum value respectively. Box starts from the first quartile to the third quartile. A vertical line goes through the box at the median.

From the given box plot we can conclude that

For Team A:

$Minimum = 80,Q_1=87, median=91, Q_3=95,Maximum=110$

For Team B:

$Minimum = 79,Q_1=82, median=87, Q_3=91,Maximum=103$

Formula for interquartile range.

$IQR=Q_3-Q_1$

1.

IQR of Team A.

$IQR_{A}=95-87=8$

Option C is correct.

2.

IQR of Team B.

$IQR_{B}=91-82=9$

Option D is correct.

3.

Difference of the medians of team A and team B is

$91-87=4$

Option A is correct.

4.

We know that 4 is about half of 8 and 9.

It means 4 is about half the interquartile range of either data set.

Option A is correct.

Paraphin [41]2 years ago

6 0

1. 95-87 = 8 /2 = 4

2. 91 - 82 = 9 /2 = 4.5 =5

3. 91 - 87 = 4

4. B 1 times

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Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

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