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Elis [28]
2 years ago
7

Kindly answer the first question . LCM of 20,35 and 30​

Mathematics
1 answer:
lukranit [14]2 years ago
3 0

<em>R</em><em>e</em><em>f</em><em>e</em><em>r</em><em> </em><em>t</em><em>o</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>a</em><em>t</em><em>t</em><em>a</em><em>c</em><em>h</em><em>m</em><em>e</em><em>n</em><em>t</em><em> </em><em>f</em><em>o</em><em>r</em><em> </em><em>s</em><em>o</em><em>u</em><em>l</em><em>t</em><em>i</em><em>o</em><em>n</em><em>!</em><em>!</em><em> </em>:)

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What the min, max, lower quartile, upper quartile, and median on the photo.
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Answer:

Minimum: 14 Maximum: 87.5 Lower Quartile: 34  Upper Quartile: 65  Median: 47.5

Step-by-step explanation:

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3 years ago
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RST is 124
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Solve for 4(3+x) = 8​
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3 years ago
Read 2 more answers
You are presented with three urns. each urn has 10 marbles. in addition, each urn holds only black or red marbles. urn 1 has 3 r
Pachacha [2.7K]
To calculate this probability we must take into account that there is the same probability that any of the 3 urns is chosen.
 This probability is:
 P (U1) = P (U2) = P (U3) = 1/3
 Urn 1 contains 7 black and 3 red marbles
 Urn 2 contains 2 black and 8 marbles network
 Urn 3 contains 5 black marbles and 5 red marbles.

 The probability of obtaining a black marble in Urn 1 is 7/10.
 The probability of obtaining a black marble in Urn 2 is 2/10
 The probability of obtaining a black marble in Urn 3 is 5/10.

 Then we look for the probability of obtaining a black marble from urn 1 or a black marble from urn 2 or a black marble from urn 3. This is:
 P (U1yB) + P (U2yB) + P (U3yB)
 So:
 (1/3) * (7/10) + (1/3) * (2/10) + (1/3) * (5/10) = 0,2333 + 0,0667 + 0,1667 = 0, 4667.
 The probability that it is a black marble is 46.67%
5 0
3 years ago
1) The Harrison family traveled 2112 mile in 4 days on their last vacation. The family traveled the same distance each day. How
oee [108]
\frac{2112}{4}\\528

1. c

2. j

\frac{30}{1\frac{1}{4}}\\\frac{30}{\frac{5}{4}}\frac{120}{5}\\24

3. a

0.03*36.8\\1.104

4. g

\frac{24\frac{1}{2}}{1\frac{3}{4}}\\\frac{\frac{49}{2}}{\frac{7}{4}}\\\frac{98}{7}\\14

5. g
3 0
3 years ago
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