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Aleksandr [31]
3 years ago
6

A letter is selected from the words in accommodation list all the possible outcomes

Mathematics
1 answer:
Luda [366]3 years ago
8 0
Since a letter is selected from      a c c o m m o d a t i o n

Possible outcome:

It could be any of the thirteen letters. There are thirteen possibilities.

a      c     c     o       m     m     o     d     a    t      i     o     n          
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A window in the shape of a parallelogram has the dimensions given.
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A. 20ft

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Through: (1, -3), slope = -4<br> A) y=-x+4 B) y = x +4<br> C) y = 4x + 1 D) y=-4x + 1
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Round 14.857 to the nearest tenth. A. 15.0 B. 14.8 C. 14.9 D. 10.0
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The height of Mountain P is 1,086 feet.The height of Mountain Q is 4 times the height of Mountain P.The area model shown below r
earnstyle [38]

Answer:

A=4000, B=80, C=24

Step-by-step explanation:

You forgot to put the correct area model, I attached it to the answer.

We have the fact that Mountain Q is 4 times the height of Mountain P. That's the "4" we have in the left side of our model. It's like having a multiplication table, next to the "4" we have "A" and upper this we have "1000", the only thing we have to do is multiplify 4*1000=4000. The next letter we have is B and below it we have "320", we divided it by 4, 320/4=80. The last letter we have is C, and is below a "6", we only have to multiplify it by 4, 6*4=24.

At the end we only sum our

  • A + 320 + c = 4344 (4 times the height of Mountain P).
  • 1000 + B + 6 = 1086(the height of the Mountain P).

8 0
3 years ago
For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12
notsponge [240]
The binomial distribution is given by, 
P(X=x) =  (^{n}C_{x})p^{x} q^{n-x}
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability <span>of obtaining a score less than or equal to 12.
</span>∴ P(X≤12) = (^{100}C_{x})(0.2)^{x} (0.8)^{100-x}
                    where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12                  
∴ P(X≤12) = (^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1} + (^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3} + (^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5} + (^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7} + (^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9} + (^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11} + (^{100}C_{12})(0.2)^{12} (0.8)^{100-12}


Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability. 
7 0
3 years ago
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