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CaHeK987 [17]
2 years ago
5

A bag has 4 red marbles, 5 blue marbles, and 6 green marbles. What is the probability if choosing a red marble from the bag, not

putting it back, and then choosing a green marble?
Mathematics
1 answer:
raketka [301]2 years ago
6 0

Answer: 4/35

Step-by-step explanation:

Number of red marbles = 4

Number of blue marbles = 5

Number of green marbles = 6

Total number of marbles = 4+5+6 = 15

When a red marble is picked, the probability is 4/15, then to pick a green marble after that, we have 14 marbles left and probability of picking green is 6/14.

Now, probability of choosing a red marble from the bag, not putting it back, and then choosing a green marble will be:

= 4/15 × 6/14

= 24/210

= 4/35

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Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

C=2\pi r

<u><em>Find the radius of the sphere with a 5.50-meter circumference</em></u>

For C=5.50\ m

assume

\pi =3.14

substitute and solve for r

5.50=2(3.14)r

r=5.50/[2(3.14)]=0.88\ m

<u><em>Find the radius of the sphere with a 7.85-meter circumference</em></u>

For C=7.85\ m

assume

\pi =3.14

substitute and solve for r

7.85=2(3.14)r

r=7.85/[2(3.14)]=1.25\ m

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

SA=4\pi r^{2}

<u><em>Find the surface area of sphere with a 5.50-meter circumference</em></u>

For r=0.88\ m

assume

\pi =3.14

substitute

SA=4(3.14)(0.88)^{2}

SA=9.73\ m^{2}

<u><em>Find the surface area of sphere with a 7.85-meter circumference</em></u>

For r=1.25\ m

assume

\pi =3.14

substitute

SA=4(3.14)(1.25)^{2}

SA=19.63\ m^{2}

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

<u><em>Find the cost of sphere with a 5.50-meter circumference</em></u>

9.73*(92)=\$895.16

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

<u><em>Find the cost of sphere with a 7.85-meter circumference</em></u>

19.63*(92)=\$1,805.96

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

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3 years ago
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