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2 years ago

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If each coded item in a catalog begins with 3 distinct letters followed by 4 distinct nonzero digits, ﬁnd the probability of ran

domly selecting one of these coded items with the ﬁrst letter a vowel and the last digit even.

1 answer:

laila [671]2 years ago

3 0

Answer: 10/117

Step-by-step explanation:

3 distinct letters and 4 distinct nonzero digits

For letters: A → Z = 26 options

For numbers: 1 → 9 = 9 options

As we have 26 options for letters and 9 for numbers and they have to be distinct, for the total of codes we have

Total of codes: 26.25.24.9.8.7.6

For the particular code asked, we have 5 options of vowels and 4 options of even numbers, so

5.25.24.8.7.6.4

probability: <u>5.25.24.8.7.6.4</u> = <u>10</u>

26.25.24.9.8.7.6 117

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A sailboat is sitting at rest near its dock. A rope attached to the bow of the boat is drawn in over a pulley that stands on a p

irina [24]

Answer:

The boat is approaching the dock at rate of 2.14 ft/s

Step-by-step explanation:

The situation given in the question can be modeled as a triangle, please refer to the attached diagram.

A rope attached to the bow of the boat is drawn in over a pulley that stands on a post on the end of the dock that is 5 feet higher than the bow that means x = 5 ft.

The length of rope from bow to pulley is 13 feet that means y = 13 ft.

We know that Pythagorean theorem is given by

$x^{2} + y^{2} = z^{2}$

Differentiating the above equation with respect to time yields,

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}$

$x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$

dx/dt = 0 since dock height doesn't change

$y\frac{dy}{dt} = z\frac{dz}{dt}$

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

The rope is being pulled in at a rate of 2 feet per second that is dz/dt = 2 ft/s

First we need to find z

z² = (5)² + (13)²

z² = 194

z = √194

z = 13.93 ft

So,

$\frac{dy}{dt} = \frac{z}{y} \frac{dz}{dt}$

$\frac{dy}{dt} = \frac{13.93}{13}(2)$

$\frac{dy}{dt} = 2.14$ $ft/s$

Therefore, the boat is approaching the dock at rate of 2.14 ft/s

8 0

1 year ago

Lois says any addition equation where the addends are all the same can be written as a multiplication equation. Is Lois correct

USPshnik [31]

Lois is correct. This is because multiplication is repeated addition. It means that when your addends are just the same number but added several times, the sum is just equivalent to the product of that number times the number of times it was added. For example, 4+4+4 = 4×3 = 12.

6 0

1 year ago

Computer upgrades have a nominal time of 80 minutes. samples of five observations each have been taken, and the results are as l

Goshia [24]

The answer:
<span>the upper and lower control limits (uclim and lclim) for mean formula is

for the mean chart
uclim= x+A2xR
where x = sum(of the value) / number of each value
and for
lclim=</span>x+A2xR
<span>
R is the range such that R= Xmax - Xmin

in the case of the sample 1: S1
the data are:
79.2 78.8 80.0 78.4 81.0

the mean is x1 = (</span>79.2 + 78.8 + 80.0 + 78.4 + 81.0) / 5= 79.48
<span>its range is R 1= 81.0 -78.4 = 2.6

we can do the same method for finding the mean chart and range for all samples
</span>S2: x2=<span> 80.14 , R2=2.3
</span>S3: x3= 80.14 , R3=1.2
S4: x4= 79.60 , R4=1.7
S5: x5= 80.02 , R5=2.0
S6: x6=80.38 , R6=1.4
<span>
therefore the average value is X= sum( x1+x2+...+x6) / 6 = 79.96
and R=sum(R1+R2+...+R6)/6=1.87
finally
range chart uclim =D4xR=3.95 and lclim is always equal to 0, because D3=0

we can say that the process is not in control.

</span>

3 0

2 years ago

Given: circle O with tangent AB mBC= 2x -16; mCD = x+40; mDE = x; mEB = 60 <br><br>find the x

murzikaleks [220]

Answer:

69

Step-by-step explanation:

All the given arcs cover the entire circle circumference, so their measures add up to a full 360.

(2x - 16) + (x + 40) + x + 60 = 360

4x + 84 = 360

4x = 276

x = 69

6 0

1 year ago

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Find the lengths of the median o

nexus9112 [7]

Answer:

$5/2units\\$ .

$\sqrt{47/2}units \\$ .

$\sqrt{41/2}units\\$ .

Step-by-step explanation:

The given points are A(1,2,3), B(-2,0,5) and C(4,1,5). The triangle is represented in the attach file where the three possible median are length AE, BF, and CD. We determine the coordinate of point D,E and F using the midpoint equation which is for any point A(x,y,z) and point B(a,b,c), the midpoint D is determine by

$D=(\frac{x+a}{2},\frac{y+b}{2},\frac{z+c}{2})\\$ .

Hence going by the above formula we determine the coordinate of point D,E and F

$D=(\frac{-2+1}{2},\frac{0+2}{2},\frac{5+3}{2})\\$ .

$D=(\frac{-1}{2},1,4)\\$ .

point E

$E=(\frac{4-2}{2},\frac{1+0}{2},\frac{5+5}{2})\\$ .

$E=(1,\frac{1}{2},5)\\$ .

Point F

$F=(\frac{4+1}{2},\frac{1+2}{2},\frac{5+3}{2})\\$ .

$F=(\frac{5}{2},\frac{3}{2},4)\\$ .

To determine the length of each median line we use the formula for distance between two points which is express as

$AB=\sqrt{(y_{2}-y_{1} )^{2} +(x_{2}-x_{1} )^{2}+(z_{2}-z_{1} )^{2}} \\$ .

Using the above formula we determine the length of line AE,BF and CD.

$AE=\sqrt{(1-1 )^{2} +(1/2-2)^{2}+(5-3 )^{2}} \\$ .

$AE=\sqrt{0 +9/4+4} \\$ .

$AE=\sqrt{25/4} \\$ .

$AE=5/2units\\$ .

For point BF

$BF=\sqrt{(5/2+2 )^{2} +(3/2-0)^{2}+(4-5 )^{2}}\\$ .

$BF=\sqrt{81/4 +9/4+1} \\$ .

$BF=\sqrt{47/2} \\$ .

$BF=\sqrt{47/2}units \\$ .

For point CD

$CD=\sqrt{(-1/2-4 )^{2} +(1-1)^{2}+(4-5 )^{2}}\\$ .

$BF=\sqrt{81/4 +0+1} \\$ .

$BF=\sqrt{41/2} \\$ .

$BF=\sqrt{41/2}units\\$ .

7 0

1 year ago