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Alja [10]
4 years ago
12

Help me pleaseeee Find x and y.

Mathematics
2 answers:
Alex777 [14]4 years ago
7 0

Answer:

x = 45°

y = 15.41 units

Step-by-step explanation:

Given quadrilateral is a square. Each angle of square is 90°

Diagonal of square bisects the angle.

\therefore \: x =  \frac{90 \degree}{2}  = 45 \degree \\  \\  \because \: diagonal \: of \: square \: is \:  \sqrt{2}  \: times \: the \:  \\ side \: of \: square \\  \therefore \: y =  \sqrt{2}  \times 10.9 \\ \therefore \: y =  1.414  \times 10.9 \\ \therefore \: y  = 15.4126 \\ \therefore \: y  \approx \: 15.41 \: units

MariettaO [177]4 years ago
3 0

Answer:

x = 45º and y = 15.414927836…

Step-by-step explanation:

To find xº, we can tell that every angle of the square is 90º. If xº is ½ of one of those angles, x = 90 ÷2 = 45º. We can find y by using the Pythagorean Theorem, which consists of the formula a² + b² = c². 10.9² + 10.9² = 237.62. <em>y </em>is the square root of 237.62. √237.62 = 15.41492783… So we can tell x = 45º and y = 15.41492783…

Hope this helps! Plz give me brainliest, it will help me achieve my next rank.

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a) The probability that exactly 17 of them enroll in college is 0.116.

b) The probability that more than 14 enroll in college is 0.995.

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P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{29}{k} 0.65^{k} 0.35^{29-k}\\\\\\

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b) The probability that more than 14 of them enroll in college is:

P(X>14)=\sum_{15}^{29} P(X=k_i)=1-\sum_{0}^{14} P(X=k_i)\\\\\\P(x=0)=0\\\\P(x=1)=0\\\\P(x=2)=0\\\\P(x=3)=0\\\\P(x=4)=0\\\\P(x=5)=0\\\\P(x=6)=0\\\\P(x=7)=0\\\\P(x=8)=0\\\\P(x=9)=0\\\\P(x=10)=0.001\\\\P(x=11)=0.002\\\\P(x=12)=0.005\\\\P(x=13)=0.013\\\\P(x=14)=0.027\\\\\\P(X>14)=1-0.005=0.995

c) Using the probabilities calculated in the point b, we  have:

P(X

d) The probabilities that more than 24 enroll in college is:

P(X>24)=\sum_{25}^{29}P(X=k_i)\\\\\\ P(x=25) = \dbinom{29}{25} p^{25}(1-p)^{4}=23751*0*0.015=0.007\\\\\\P(x=26) = \dbinom{29}{26} p^{26}(1-p)^{3}=3654*0*0.043=0.002\\\\\\P(x=27) = \dbinom{29}{27} p^{27}(1-p)^{2}=406*0*0.123=0\\\\\\P(x=28) = \dbinom{29}{28} p^{28}(1-p)^{1}=29*0*0.35=0\\\\\\P(x=29) = \dbinom{29}{29} p^{29}(1-p)^{0}=1*0*1=0\\\\\\\\P(X>24)=0.007+0.002+0+0+0=0.009

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