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3 years ago

Which statement is true?

Mathematics

1 answer:

olga_2 [115]3 years ago

6 0

Answer: A. Paul’s data has a bigger overall spread than Sally’s data

The range of Paul’s Data: 28-12=16
The range of Sally’s Data: 30-15=15
16>15. Therefore Paul’s data has a bigger overall spread than Sally’s

Why B is wrong: Remember that the interquartile range is the “box” range.
Paul’s IQR: 26-15=11
Sally’s IQR: 28-18=10
11<10 is false, so B is false.

Why C and D is wrong: The middle line in the IQR represented in each box and whisker plot is the median. Paul’s median is not equal nor greater than Sally’s median. 18=24 18<24 is false. Therefore, C and D are false.

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Suppose you have a litter of mice which consists of 8 males and 4 females. If you randomly grab two of them, what is the probabi

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The probability that they are both male is 0.424 (3 d.p.)

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Step-by-step explanation:

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A metal beam was brought from the outside cold into a machine shop where the temperature was held at 70°F After 10 min, the beam

blondinia [14]

Answer:

The beam's initial temperature was 10°F

Step-by-step explanation:

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature. This means that:

$\frac{dT}{dt} =-k (T-T_{a})$ where k is a positive constant and $T_{a}$ is the ambient temperature.

This is the solution of the differential equation

$T(t)=T_{a}+T_{0}\cdot e^{(kt)}$ where $T(t)$ is the temperature after t minutes and $T_{0}$ and k are constants yet to be determined.

We know from the information given that the ambient temperature is 70°F, so

$T(t)=70+T_{0}\cdot e^{(kt)}$

We also know that $T(10) = 40 \:F$ and $T(20) = 55 \:F$ , we can use these to determine the constants $T_{0}$ and k.

If we use the first condition $T(10) = 40 \:F$ we have

$40=70+T_{0}\cdot e^{(k\cdot 10)}$

We can solve for k in terms of $T_{0}$ as follows

$40=70+T_{0}\cdot e^{(k\cdot 10)}\\70+T_0e^{k\cdot 10}=40\\T_0e^{k \cdot 10}=-30\\e^{k \cdot 10}=-\frac{30}{T_0}\\\ln \left(e^{k\cdot \:10}\right)=\ln \left(-\frac{30}{T_0}\right)\\k\cdot \:10\ln \left(e\right)=\ln \left(-\frac{30}{T_0}\right)\\k=\frac{\ln \left(-\frac{30}{T_0}\right)}{10}$

We can rewrite $T(t)$ as

$T(t)=70+T_{0}\cdot e^{(\frac{\ln \left(-\frac{30}{T_0}\right)}{10}\cdot t)}$

Next we use the second condition $T(20) = 55 \:F$ to get

$55=70+T_{0}\cdot e^{(\frac{\ln \left(-\frac{30}{T_0}\right)}{10}\cdot 20)}$

and we solve for $T_{0}$

$55=70+T_{0}\cdot e^{(\frac{\ln \left(-\frac{30}{T_0}\right)}{10}\cdot 20)}\\-15=T_{0}\cdot e^{2\ln \left(-\frac{30}{T_0}\right)}\\-15=T_{0}\cdot e^{\ln \left(-\frac{30}{T_0}\right)^{2}}\\-15=T_{0}\cdot \left(-\frac{30}{T_0}\right)^{2}\\-15=T_{0} \cdot \left(\frac{900}{T_0^2}\right)\\-15=\frac{900}{T_{0}} \\T_{0} = -60$

The value of k is

$k=\frac{\ln \left(-\frac{30}{T_0}\right)}{10}\\k=\frac{\ln \left(\frac{-30}{-60}\right)}{10}\\k=-\frac{ln(2)}{10}$

So the general solution of the equation is

$T(t)=70-60\cdot e^{(-\frac{ln(2)}{10}\cdot t)}$

In particular, since we want to know T(0), we can now just evaluate:

$T(0)=70-60\cdot e^{(-\frac{ln(2)}{10}\cdot 0)}\\T(0)=10$

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