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3 years ago

Is a triangle with sides that measure 6 inches, 8 inches, and 9 inches a right triangle?

2 answers:

6 0

It is a
r
i
g
h
t
−
∠
e
d
Δ
, as it has got one angle of measure
90
∘
.

Explanation:
Even if it had not been mentioned that the
Δ
has one
90
∘
∠
, we
can infer this fact from the given lengths of the sides.
Observe that,
6
2
+
8
2
=
10
2
; hence, by the converse of
Pythagoras' Theorem, the
∠
opp. to side with length
10
'
is a
right-angle.
Enjoy Maths.!

frez [133]3 years ago

3 0

Answer:

Step-by-step explanation:

A right triangle needs to have a right angle in this case this triangle does not. The next answer up is a triangle with 6,8, and 10 inches. This would give you a right triangle.

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When network cards are communicating, bits can occasionally be corrupted in transmission. Engineers have de- termined that the n

ycow [4]

Answer:

a) 11.40% probability of 5 bits being in error during the transmission of 1 kb

b) 11.60% probability of 8 bits being in error during the transmission of 2 kb

c) 0.01% probability of no error bits in 3kb

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the giveninterval.

Poisson distribution with mean of 3.2 bits/kb (per kilobyte).

This means that $\mu = 3.2kb$ , in which kb is the number of kilobytes.

(a) What is the probability of 5 bits being in error during the transmission of 1 kb?

This is P(X = 5) when $\mu = 3.2$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 5) = \frac{e^{-3.2}*3.2^{5}}{(5)!} = 0.1140$

11.40% probability of 5 bits being in error during the transmission of 1 kb

(b) What is the probability of 8 bits being in error during the transmission of 2 kb?

This is P(X = 8) when $\mu = 2*3.2 = 6.4$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 8) = \frac{e^{-6.4}*6.4^{8}}{(8)!} = 0.1160$

11.60% probability of 8 bits being in error during the transmission of 2 kb

(c) What is the probability of no error bits in 3kb?

This is P(X = 0) when $\mu = 3*3.2 = 9.6$

Then

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-9.6}*9.6^{0}}{(0)!} = 0.0001$

0.01% probability of no error bits in 3kb

7 0

3 years ago

What is the volume of the cone?

Lady bird [3.3K]

Answer: C: 2787.64
Explanation: The height of the cone equals 22, the diameter of the circle is 22, divide it by 2 to get the radius which is 11. Then plug in data into formula: V=

Pi•radius squared• height
____________________

3

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4 years ago

I need to know the answers please

Rus_ich [418]

It is 180 - 65

Which is 115 which would be your answer

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3 years ago

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What is the difference between -4 and 6?

rusak2 [61]

2 is the difference

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