What is the mean absolute deviation of the data set? {12, 4, 30, 16, 18} 2. 8 5 2006. 4 16.

If sum of two no is 21 and sum of their square is 256 find the number

polet [3.4K]

Step-by-step explanation:

you're going to solve for y in the first equation and then substitute it in for the second equation and solve.

4 0

3 years ago

Carpet costs $7.84 per yard. What is the cost of 21.5 yards?

balandron [24]

Answer:

$168.56

Step-by-step explanation:

8 0

3 years ago

According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildif

suter [353]

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

$\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018$

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

$z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55$

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

$P(0.79$

3 0

3 years ago

Write an equation of the line that passes through the given points.<br> (-2,5) and (1,2)

Anon25 [30]

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{(-2)}}}\implies \cfrac{-3}{1+2}\implies \cfrac{-3}{3}\implies -1$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{-1}[x-\stackrel{x_1}{(-2)}] \\\\\\ y-5=-(x+2)\implies y-5=-x-2\implies y=-x+3$