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zmey [24]
3 years ago
12

Thomas bought a bag of jelly beans that contained 10 red jelly beans, 15 blue jelly beans, and 12 green jelly beans. What is the

probability of Thomas reaching into the bag and pulling out a blue or green jelly bean and then reaching in again and pulling out a red jelly bean? Assume that the first jelly bean is not replaced.
A. 1/100
B. 3
C. 5/74
D. 0
Mathematics
1 answer:
Hitman42 [59]3 years ago
7 0
There are 37 jelly beans so p for the first one is 27/37. That leaves 36 in the bag. So p for the second jelly bean is 10/36=5/18.
The combine probability is 27/37×5/18=3/37×5/2=15/74.
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<span>6840 customers ... 45 days<span>
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4 0
3 years ago
6-10 divide, another onee thank you!!​
eduard

Answers:

10.) \displaystyle \pm{5}

9.) \displaystyle 1\frac{1}{2}

8.) \displaystyle \pm{1\frac{1}{2}}

7.) \displaystyle \pm{1\frac{1}{2}}

6.) \displaystyle \pm{\frac{1}{2}}

Step-by-step explanations:

10.) \displaystyle \frac{\sqrt{200}}{\sqrt{8}} \hookrightarrow \sqrt{25} \hookrightarrow \frac{\pm{10\sqrt{2}}}{\pm{2\sqrt{2}}} \\ \\ \boxed{\pm{5}}

9.) \displaystyle \frac{\sqrt[3]{135}}{\sqrt[3]{40}} \hookrightarrow \sqrt[3]{3\frac{3}{8}} \hookrightarrow \frac{3\sqrt[3]{5}}{2\sqrt[3]{5}} \\ \\ \boxed{1\frac{1}{2}}

8.) \displaystyle \frac{\sqrt[4]{162}}{\sqrt[4]{32}} \hookrightarrow \sqrt[4]{5\frac{1}{16}} \hookrightarrow \frac{\pm{3\sqrt[4]{2}}}{\pm{2\sqrt[4]{2}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}

7.) \displaystyle \frac{\sqrt{63}}{\sqrt{28}} \hookrightarrow \sqrt{2\frac{1}{4}} \hookrightarrow \frac{\pm{3\sqrt{7}}}{\pm{2\sqrt{7}}} \\ \\ \boxed{\pm{1\frac{1}{2}}}

6.) \displaystyle \frac{\sqrt{12}}{\sqrt{48}} \hookrightarrow \sqrt{\frac{1}{4}} \hookrightarrow \frac{\pm{2\sqrt{3}}}{\pm{4\sqrt{3}}} \\ \\ \boxed{\pm{\frac{1}{2}}}

I am joyous to assist you at any time.

5 0
2 years ago
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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Time in hours = 2800/(60*60)

Time in hours= 2800/360

Time in hours= 0.778 hours

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