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liubo4ka [24]
3 years ago
8

Can someone please help me with this one??

Mathematics
1 answer:
GrogVix [38]3 years ago
8 0

Answer:

x = 3.6 cm

Step-by-step explanation:

By the theorem of intersecting secants,

"If two secants are drawn from a point outside the circle, product of the lengths of the secant segment and its external segment are equal."

3(3 + y) = 2(2 + 6 + 3)

9 + 3y = 2 × 11

3y = 22 - 9

3y = 13

y = \frac{13}{3} cm = 4.33 cm

Now we will apply theorem of intersecting chords to determine the value of x.

" When two chords intersect each other in a circle, product of their segments are equal"

x\times 5=6\times 3

x=\frac{18}{5}

x=3.6 cm

Therefore, x = 3.6 cm and y = 4.33 cm

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a. p=1.000

b. p=0.2924

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Step-by-step explanation:

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P(X=x){n\choose x}p^x(1-p)^{n-x}

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Hence, the probability that all 12 flights are on time is 1.0000

b. Given that n=12, and p=0.85

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P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\={12\choose 10}0.85^{10}(1-0.85)^{2}\\\\=0.2924

Hence, the probability that exactly 10 flights are on time is 0.2924

c. Given that n=12, and p=0.85

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P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\P(X\geq 10)=P(X=10)+P(X=11)+P(X=12)\\\\={12\choose 10}0.85^{10}(0.15)^{2}+{12\choose 11}0.85^{11}(0.15)^{1}+{12\choose 12}0.85^{12}(0.15)^{0}\\\\=0.2924+0.3012+0.1422\\\\=0.7358

Hence, the probability of 10+ flights being on time is 0.7358

d. We first find the mean of the distribution:

\mu=E(X)=0.85\times 14\\\\=11.9

#We then find the probability of 11+=0.3012+0.1422=0.4434

-We compare the expectation to the probability of 11+ flights being on time.

No. Since the probability P(X\geq 10)=0.4434 < that the expectation, 11.9, it is not unusual  for 11+ flights to be on time.

*I have used a sample size of n=12 since there are two separate n values:

5 0
3 years ago
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Answer = $55
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