Answer:
<h2>1.84feet</h2>
Step-by-step explanation:
Using the formula for finding range in projectile, Since range is the distance covered in the horizontal direction;
Range ![R = U\sqrt{\frac{H}{g} }](https://tex.z-dn.net/?f=R%20%3D%20U%5Csqrt%7B%5Cfrac%7BH%7D%7Bg%7D%20%7D)
U is the velocity of the arrow
H is the maximum height reached = distance below the bullseye reached by the arrow.
R is the horizontal distance covered i.e the distance of the target from the archer.
g is the acceleration due to gravity.
Given R = 60ft, U = 250ft/s, g = 32ft/s H = ?
On substitution,
![60 = 250\sqrt{\frac{H}{32}} \\\frac{60}{250} = \sqrt{\frac{H}{32}}\\\frac{6}{25} = \sqrt{\frac{H}{32}](https://tex.z-dn.net/?f=60%20%3D%20250%5Csqrt%7B%5Cfrac%7BH%7D%7B32%7D%7D%20%5C%5C%5Cfrac%7B60%7D%7B250%7D%20%3D%20%5Csqrt%7B%5Cfrac%7BH%7D%7B32%7D%7D%5C%5C%5Cfrac%7B6%7D%7B25%7D%20%3D%20%5Csqrt%7B%5Cfrac%7BH%7D%7B32%7D)
Squaring both sides we have;
![(\frac{6}{25} )^{2} = (\sqrt{\frac{H}{32} } )^{2} \\\frac{36}{625} = \frac{H}{32} \\625H = 36*32\\H = \frac{36*32}{625} \\H = 1.84feet](https://tex.z-dn.net/?f=%28%5Cfrac%7B6%7D%7B25%7D%20%29%5E%7B2%7D%20%3D%20%28%5Csqrt%7B%5Cfrac%7BH%7D%7B32%7D%20%7D%20%29%5E%7B2%7D%20%5C%5C%5Cfrac%7B36%7D%7B625%7D%20%3D%20%20%5Cfrac%7BH%7D%7B32%7D%20%5C%5C625H%20%3D%2036%2A32%5C%5CH%20%3D%20%5Cfrac%7B36%2A32%7D%7B625%7D%20%5C%5CH%20%3D%201.84feet)
The arrow will hit the target 1.84feet below the bullseye.
Answer:
90 cents
Step-by-step explanation
divide .99 by 10 and that gets you 9 cents minus that off and its 90 Cents
Answer:
![z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B0.872-0.917%7D%7B%5Cfrac%7B0.303%7D%7B%5Csqrt%7B37%7D%7D%7D%3D%20-0.903)
And we can find this probability to find the answer:
![P(z](https://tex.z-dn.net/?f=%20P%28z%3C-0.903%29)
And using the normal standar table or excel we got:
![P(z](https://tex.z-dn.net/?f=%20%20P%28z%3C-0.903%29%3D0.1833%20)
Step-by-step explanation:
Let X the random variable that represent the amounts of nocatine of a population, and for this case we know the distribution for X is given by:
Where
and
We have the following info from a sample of n =37:
the sample mean
And we want to find the following probability:
![P(\bar X \leq 0.872)](https://tex.z-dn.net/?f=%20P%28%5Cbar%20X%20%5Cleq%200.872%29)
And we can use the z score formula given by;
![z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D)
And if we find the z score for the value of 0.872 we got:
![z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B0.872-0.917%7D%7B%5Cfrac%7B0.303%7D%7B%5Csqrt%7B37%7D%7D%7D%3D%20-0.903)
And we can find this probability to find the answer:
![P(z](https://tex.z-dn.net/?f=%20P%28z%3C-0.903%29)
And using the normal standar table or excel we got:
![P(z](https://tex.z-dn.net/?f=%20%20P%28z%3C-0.903%29%3D0.1833%20)
32*1.2=38.4 (.2 is the 20%, and the 1 adds the percent to the original cost making it a mark up)
Answer:
525 x 1,050
A = 551,250 m²
Step-by-step explanation:
Let 'L' be the length parallel to the river and 'S' be the length of each of the other two sides.
The length of the three sides is given by:
![2S+L=2,100\\ L=2100-2S](https://tex.z-dn.net/?f=2S%2BL%3D2%2C100%5C%5C%20L%3D2100-2S)
The area of the rectangular plot is given by:
![A=S*L\\A=S(2100-2S)\\A=2100 S -2S^2](https://tex.z-dn.net/?f=A%3DS%2AL%5C%5CA%3DS%282100-2S%29%5C%5CA%3D2100%20S%20-2S%5E2)
The value of 'S' for which the area's derivate is zero, yields the maximum total area:
![\frac{dA(S)}{dS}=\frac{d(2100 S -2S^2)}{dS}\\0= 2100 - 4S\\S=525](https://tex.z-dn.net/?f=%5Cfrac%7BdA%28S%29%7D%7BdS%7D%3D%5Cfrac%7Bd%282100%20S%20-2S%5E2%29%7D%7BdS%7D%5C%5C0%3D%202100%20-%204S%5C%5CS%3D525)
Solving for 'L':
![L=2100-(2*525)\\L=1,050](https://tex.z-dn.net/?f=L%3D2100-%282%2A525%29%5C%5CL%3D1%2C050)
The largest area enclosed is given by dimension of 525 m x 1,050 and is:
![A = 525*1,025\\A=551,250\ m^2](https://tex.z-dn.net/?f=A%20%3D%20525%2A1%2C025%5C%5CA%3D551%2C250%5C%20m%5E2)