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gtnhenbr [62]
3 years ago
15

A bag of marbles has 5 green, 2 yellow, 8 red, and 3 purple marbles. What is the probability of picking a red marble, not replac

ing it, and then picking another red marble?
Mathematics
1 answer:
hram777 [196]3 years ago
7 0
Sample space ={5G, 2Y, 8R, 3P} =18 POSSIBLE OUTCOME

1st draw: P(R) = 8/18 = 4/9 = 0.444
2nd draw (no replacement, means already one Red is picked up, so the sample space has been reduced): P(another one more RED) = 7/17 =0.411 
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45 points for first most detailed / accurate answer:
Bas_tet [7]
A.) Since there are no restrictions as to the dimensions of the candle except that their volumes must equal 1 cubic foot and that each must be a cylinder, we have the freedom to decide the candles' dimensions. 

I decided to have the candles equal in volume. So, 1 cubic foot divided by 8 gives us 0.125 cubic foot, 216 in cubic inches.

With each candle having a volume of 216 cubic inches, I assign a radius to each: 0.5 in, 1.0 in, 1.5 in, 2.0 in, 2.5 in, 3.0 in, 3.5 in, and 4.0 in. Then, using the formula of the volume of a cylinder, which is:

V=pi(r^2)(h)

we then solve the corresponding height per candle. Let us let the value of pi be 3.14.

Hence, we will have the following heights (expressed to the nearest hundredths) for each of the radius: for 

r=2.5 in: h=11.01 in
r=3.0 in: h= 7.64 in
r=3.5 in: h= 5.62 in
r=4.0 in: h= 4.30 in
r=4.5 in: h= 3.40 in
r=5.0 in: h= 2.75 in
r=5.5 in: h= 2.27 in
r=6.0 in: h= 1.91 in

b. each candle should sell for $15.00 each

($20+$100)/8=$15.00

c. yes, because the candles are priced according to the volume of wax used to make them, which in this case, is just the same for all sizes
7 0
3 years ago
Is 312mg greater than 312dg
adelina 88 [10]

312 mg is not greater than 312 dg

5 0
3 years ago
An arc of a circle is 3πm long, and it subtends an angle of 72° at the center of the circle. Find the radius of the circle.
Alex777 [14]
72⁰_______3π m
360⁰______15π m

C = 15π = 2π r
r = 7•5 m
8 0
3 years ago
Round the following dollar amount to the nearest cent.<br><br> $5.8222
xz_007 [3.2K]

Answer:

$5.82

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Directions: Complete all 3 questions. Make sure you include a graph, work and conclusion.
Simora [160]

Prove that the quadrilateral whose vertices are I(-2,3), J(2,6), K(7,6), and L(3, 3) is a rhombus.

I think in these problems the first step is to express each side as a vector.  A vector is the difference between points.  When two sides have the same vector (or negatives) it means they're parallel and congruent.  So in a rhombus IJKL the vectors IJ and LK should be the same, as should JK and IL.  That much assures a parallelogram; we check IJ and JK are congruent to complete the crowing of the rhombus.

Let's calculate these vectors:

IJ = J - I = (2,6) - (-2,3) = (2 - -2, 6 - 3) = (4, 3)

LK = K - L = (7, 6) - (3, 3) = (4, 3)

IJ = LK, so far so good

(Note: If you haven't got to vectors yet you can just show the two sides are the same length, 5, and have the same slope, 3/4, both of which can be read off the vectors.)

JK = K - J = (7,6) - (2,6) = (5,0)

IL = L - I = (3, 3) - (-2, 3)  = (5, 0)

Those are the same too.    

Now we have to show IJ ≅ JK

The length of IJ is the cliche √4²+3² = 5, the same as JK, so IJ ≅ JK

We showed all four sides are congruent and we have two pair of parallel sides, so we have a rhombus.

8 0
3 years ago
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