Answer: D. The interquartile range for Forida is 11, greater than interquartile range for Australia, 4.
Step-by-step explanation:
We know that in a box-plot the vertical line between the box which divide the box into two parts represents the median of the data.
We can see that box plots are skewed .
In this case the Interquartile range is a better measure of spread than the range because the data is affected by outliers.
From the given box plots which shows the shark attack data, it can be seen that the
And
∴ The statement best compares the spreads of the data sets : The interquartile range for Forida is 11, greater than interquartile range for Australia, 4.
Answer:
The change in temperature per minute for the sample, dT/dt is 71. °C/min
Step-by-step explanation:
The given parameters of the question are;
The specific heat capacity for glass, dQ/dT = 0.18 (kcal/°C)
The heat transfer rate for 1 kg of glass at 20.0 °C, dQ/dt = 12.9 kcal/min
Given that both dQ/dT and dQ/dt are known, we have;
Therefore, we get;
For the sample, we have the change in temperature per minute, dT/dt, presented as follows;
The dimensions such as length is 18 and width is 16 or length is 32 and width is 9 will produce an enclosed area of 288ft.
<h3>What is the dimension of length and width?</h3>We know that,
P = l + 2w
50 = l + 2w
50 - 2w = l
A = l x w
288 = (50 - 2w)w
288 = 50w -2w²
2w² - 50w + 288
By dividing throughout by 2
w² - 25w + 144
This is in quadratic equation form.
By applying quadratic formula,
w = (-(-25) ± ) / 2
w = (-(-25) ± 7)/2
w1 = (-(-25) + 7)/2
w1 = (25 + 7) / 2
w1 = 16
w2 = (-(-25) - 7)/2
w2 = (25 - 7) / 2
w2 = 9
By substituting the values in eq1,
A = l x w
if w = 16
288 = l x 16
l = 288/16
l = 18
if w = 9
288 = l x 9
l = 288/9
l = 32
The dimensions will be length is 18 and width is 16 or length is 32 and width is 9.
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