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hjlf
2 months ago
13

The data set below represents the total number of touchdowns a quarterback threw each season for 10 seasons of play.

Mathematics
2 answers:
Zina [12.3K]2 months ago
7 0

The range is 24 touchdowns.

The interquartile range equals 8 touchdowns.

tester [12.3K]2 months ago
5 0
Here are the calculations for variability. The range is determined by subtracting the highest value from the lowest (29-5=24). To ascertain the interquartile range, identify the median of both the lower half and the upper half of the data then perform subtraction with these two medians. Here’s your list, with parentheses around the upper and lower quartiles: 5, 17, (18), 20, 20, 21, 23, (26), 28, 29. This is akin to finding the average of the entire dataset followed by determining the average of each half. Subtract 26 from 18 to yield an interquartile range of 8 touchdowns.
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Each day that a library book is kept past its due date, a $0.30 fee is charged at midnight. Which ordered pair is viable solutio
tester [12383]

Assume that the number of overdue days for the library book is X, and the total fine is Y.

Liability charges a fee of $0.30 per day for lateness,

Thus, for one day late, the fee equals: 1 * $0.30

For X days late, the fee becomes: X * $0.30.

The total fee Y corresponds to being X days late, so: Y = 0.30 * X.

Next, we'll verify which possible answers satisfy the equation Y = 0.30 * X.

Options one (-3, -0.9) and two (-2.5, -0.75) are invalid since the number of late days cannot be negative, and these give negative values for X.

Option three (4.5, 1.35) is also incorrect because the library charges per whole day; 4.5 days is not an integer, so they would charge 4 or 5 days, not 4.5.

Option four (8, 2.40) matches the equation perfectly:

Y = 0.30 * X

2.40 = 0.30 * 8

2.40 = 2.40.

Therefore, the only correct answer is option four (8, 2.40).

8 0
3 months ago
Helen draws a random circle.
Leona [12618]

Answer:

Upper limit: 3.116

Lower limit: 3.125

Step-by-step breakdown:

The random circle drawn by Hellen has a circumference of C=405 mm, measured to 3 significant figures, leading to a diameter, d, of 130 mm, accurate to 2 significant figures.

We utilize the formula

\pi = \frac{C}{d}

Substituting in the figures provides:

\pi = \frac{405}{130}

\pi = 3.11538461538

Consequently, Hellen’s value for π is recorded as 3.115 to three decimal places.

To find both limits, we divide the specified level of precision by two.

\frac{0.001}{2} = 0.0005

This figure is then added to the rounded number to determine the upper limit:

3.115+0.0005=3.1155

The lower limit calculates as 3.115-0.0005=3.1145

7 0
2 months ago
An airplane travels at 950 km/h. how long does it take to travel 1.00km? in hours
AnnZ [12381]
An airplane flying at a speed of 950 kilometers per hour would take 0.0010526 hours to travel 1.00 kilometer.
6 0
2 months ago
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