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1 year ago

Geometry question! Help!

Mathematics

1 answer:

Karolina [17]1 year ago

8 0

Answer: $2\sqrt{2}$

Step-by-step explanation:

The inradius is given by the area of the triangle divided by the semiperimeter.

The semiperimeter is $\frac{12+12+8}{2}=16$ .

Thus, by Heron's formula, the area is $\sqrt{16(16-12)(16-12)(16-8)}=32\sqrt{2}$ .

Therefore, the radius is $\frac{32\sqrt{2}}{16}=2\sqrt{2}}$

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HELP DUE IN 20 MINS! x = ??

DerKrebs [107]

$\huge{ \mathcal{ \underline{ Answer} \: \: ✓ }}$

We know,

$27 \times (x + 27) = 36 {}^{2}$

$x + 27 = \dfrac{1296}{27}$

$x + 27 = 48$

$x = 21$

______________________

$\mathrm{ ☠ \: TeeNForeveR \:☠ }$

4 0

2 years ago

In a home theater system, the probability that the video components need repair within 1 year is 0.02, the probability that the

earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

A = video components need repair within 1 year

B = electronic components need repair within 1 year

C = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events A, B and C are independent.

(a)

Compute the probability that at least one of these components will need repair within 1 year as follows:

P (At least 1 component needs repair)

= 1 - P (No component needs repair)

$=1-P(A^{c}\cap B^{c}\cap C^{c})\\=1-[P(A^{c})\times P(B^{c})\times P(C^{c})]\\=1-[(1-0.02)\times (1-0.007)\times (1-0.001)]\\=1-0.97216686\\=0.02783314\\\approx 0.0278$

Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

$=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277$