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Virty [35]
3 years ago
5

Can someone provide a step by step explanation on how to solve? Cheers.

Mathematics
1 answer:
blsea [12.9K]3 years ago
6 0

Answer:

a) 0.0137060668

please mark me brainliest and follow me my friend.

I not know b part.

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You are holding a kite string in your hand. The angle of elevation from your hand to the kite is 53∘and the distance to the kite
Sophie [7]

Answer:

43 degrees F.

Step-by-step explanation:

3 0
3 years ago
What is the sum of the arithmetic sequence 151, 137, 123,..if there are 26 terms?
ElenaW [278]
First term: a1 = 151
common difference: d = -14 (we decrease by 14 each time, eg, 151-14 = 137)

nth term of this arithmetic sequence is...
an = a1+d(n-1)
an = 151+(-14)(n-1)
an = 151-14n+14
an = -14n+165

This will be used in the formula below

Sn = n*(a1+an)/2
<span>Sn = n*(151+(-14n+165))/2
</span><span>S26 = 26*(151+(-14*26+165))/2 ... replace every n with 26
</span>S26 = -624

The final answer here is choice C) -624
<span>
</span>
5 0
3 years ago
A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has
bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113  

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711  

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] =  1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

7 0
3 years ago
Members of the high school baseball team need to purchase cleats before the season starts. The pitcher paid $70 for his, but the
prohojiy [21]

Answer:

The pitcher paid $70.

The second baseman paid $63.

$70 -$63 = $7

Step-by-step explanation:

6 0
3 years ago
Can someone help me
nignag [31]
The second answer, cause it best shows the equation, but they just moved the numbers around. It is also the only equation with all the same numbers used in the original equation, too
4 0
3 years ago
Read 2 more answers
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