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Nutka1998 [239]

3 years ago

11

A gun shop sells gunpowder. Monthly demand for gunpowder is normally distributed,averages 20 pounds, and has a standard deviatio

n of 2 pounds. The shop manager wishesto stock gunpowder inventory at the beginning of each month so that there is only a 2%chance that the shop will run out of gunpowder (i.e., that demand will exceed inventory)in any given month.Calculate the amount of gunpowder to stock in inventory, in pounds.
(A) 16(B) 23(C) 24(D) 32(E) 43

1 answer:

LenaWriter [7]3 years ago

8 0

Answer:

Option (C) 24

Step-by-step explanation:

Let the random variable denoting the amount of gunpowder in the shop be 'x'.

x ~ N(20, 2) i.e. (x-20)/2 ~ (0, 1)

Let 'a' be the amount (in pounds) of gunpowder to stock in inventory.

Thus, P(x<a) = 1 - 0.02 = 0.98

i.e. P[(x-20)/2<(a-20)/2] = 0.98

i.e. ф [(a-2)/2] = 0.98

i.e. (a-20)/2 = ф⁻¹ (0.98) = 2.054

i.e. a = 20 + (2 × 2.054) = 24.108 ≈ 24 pounds

24 pounds of gunpowder to stock in inventory.

Option (C) 24

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On Texas Avenue between University Drive and George Bush Drive, accidents occur according to a Poisson process at a rate of thre

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Answer:

(a) The probability is 0.6514

(b) The probability is 0.7769

Step-by-step explanation:

If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is the mean number of accidents per day and t is the number of days.

So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.

Then, the probability that a given day has no accidents is calculated as:

$P(x)=\frac{e^{-3/7}*(3/7)^{x}}{x!}$

$P(0)=\frac{e^{-3/7}*(3/7)^{0}}{0!}=0.6514$

On the other hand the probability that February has at least one accident with a personal injury is calculated as:

P(x≥1)=1 - P(0)

Where P(0) is calculated as:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:

$P(x)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{x}}{x!}$

$P(0)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{0}}{0!}=0.2231$

Finally, P(x≥1) is:

P(x≥1) = 1 - 0.2231 = 0.7769

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3 years ago

Two friends share 76 blueberries. To count the blueberries they put them

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In there mouth and they ate them

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3 years ago

16- X = -2<br> Solve for x

Licemer1 [7]

Answer: x=18

Step-by-step explanation: 16-18=-2

Good luck! :)

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2 years ago

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A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
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See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

2 years ago

Video Example EXAMPLE 1 Find the linearization of the function f(x) = x + 1 at a = 3 and use it to approximate the numbers 3.98

Blizzard [7]

Answer:

the linearization is y = 1/4x +5/4

the linearization will produce <em>overestimates</em>

the values computed from this linearization are ...

f(3.98) ≈ 2.245

f(4.05) ≈ 2.2625

Step-by-step explanation:

Apparently, you have ...

$f(x)=\sqrt{x+1}$

from which you have correctly determined that ...

$f'(x)=\dfrac{1}{2\sqrt{x+1}}$

so that f(3) = 2 and f'(3) = 1/4. Putting these values into the point-slope form of the equation of a line, we get the linearization ...

g(x) = (1/4)(x -3) +2

g(x) = (1/4)x +5/4

__

The values from this linearization will be overestimates, as the curve f(x) is concave downward everywhere. The tangent (linearization) is necessarily above the curve everywhere.

__

At the given values, we find ...

g(3.98) = 2.245

g(4.05) = 2.2625

4 0

3 years ago