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Gelneren [198K]
1 year ago
9

f(x) = sin(x), basis B = {sin(x) + cos(x), cos(x)}, basis C = {cos(x) − sin(x), sin(x) + cos(x)} in span(sin(x), cos(x))

Mathematics
1 answer:
gregori [183]1 year ago
3 0

Answer:

Step-by-step explanation:

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Find three solutions of the equation y = 7x – 5.
soldi70 [24.7K]
 y=7x-5 This is the same as saying y-7x = -5 Well easy. You just make up numbers that make this work. Let x be 1 y-7=-5 y=-5+7 y=2 So one ordered pair is (1,2) Let x=2 y-14=-5 y=9 So another ordered pair is (2,9) Let x=3 y-21=-5 y=-5+21 y=16 So another ordered pair is (3,16) 
hope it helps
5 0
2 years ago
HELPPPP i’ll give brainliest if it’s right!!!
STALIN [3.7K]

Answer:

the answer is A and D

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
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
Please someone explain this one? Thank you!
den301095 [7]
About 7,460 punds if you add everything together

5 0
3 years ago
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Triangle JKL is shown below.
Artyom0805 [142]
Where are the answer choices ?
5 0
2 years ago
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