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Vladimir79 [104]
2 years ago
8

Please help using the picture below I need the surface area I also need an explanation

Mathematics
1 answer:
Taya2010 [7]2 years ago
6 0

Answer:

155

Step-by-step explanation:

5*5=25

((5*13)/2)*4=130

130+25=155

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Round to the nearest whole number of degrees. Enter only a whole number.
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Answer:

22 degrees

Step-by-step explanation:

using inverse tangent

2.1/5.3 = 0.396226415

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ANSWER = 21.6147

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Step-by-step explanation:

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1. During a field trip, 60 students are put into equal sized groups.
loris [4]
If you use for fingers and count bye five you should eventually reach 60 and how ever many fingers you have up is your answer. Witch =7
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2 years ago
A company is allowed to interview candidates until two qualified candidates are found. But budget constraints dictate that no mo
Lapatulllka [165]

Answer:

Kindly check attached picture for sample space design

45 ways

Step-by-step explanation:

Number of qualified candidates to be chosen = 2

Number of candidates to be interviewed = 10

Combination formula :

nCr = n! / (n-r)! r!

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10C2 = 90/2

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4 0
2 years ago
A four year old is going to spin around with his arms stretched out 100 times. From past experience, his father knows it takes a
Sati [7]

Answer:

P(X \geq0.55) \leq 0.22

Step-by-step explanation:

Using central Limit Theorem (CLT), The sum of 100 random variables;

Y=X_1+X_2+...+X_{100} is approximately normally distributed with

Y ~ N (100 × \frac{1}{3^2} ) = N ( 50, \frac{100}{9} )

The approximate probability that it will take this child over 55 seconds to complete spinning can be determined as follows;

N ( 50, \frac{100}{9} )

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P(Y>55) =\phi (-1.5)

P(Y>55) =0.0668

Using Chebyshev's inequality:

P(|X-\mu\geq K)\leq \frac{\sigma^2}{K^2}

Let assume that X has a symmetric distribution:

Then:

2P(X-\mu\geq K)\leq) \frac{\sigma^2}{K^2}

2P(X \geq \mu+K)\leq) \frac{\sigma^2}{K^2}

2P(X\geq0.5+0.05)\leq \frac{\frac{1}{\frac{3^2}{100} } }{0.05^2}               where: (\sigma^2 = \frac{1}{3^2/100})

P(X \geq0.55) \leq 0.22

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3 years ago
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