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3 years ago

In triangle abc, the angle c is six times as large as angle

1 answer:

mote1985 [20]3 years ago

3 0

The interior angles of a triangle add up to 180 degrees

a = 1st angle, b = 2nd angle, c = 3rd angle

c = 6a
b = a + 60
a + b + c = 180

a + (a + 60) + (6a) = 180
8a + 60 = 180
8a = 180 - 60
8a = 120
a = 120/8
a = 15 <=== here is one angle

b = a + 60
b = 15 + 60
b = 75 <=== and another one

c = 6a
c = 6(15)
c = 90 <=== and the last angle

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What are the values of h and x in the right triangle? Explain.

alexandr402 [8]

Answer:

h = 60 ; x = 25

Step-by-step explanation:

You are going to want to use Pythagorean Theorem to solve these.

Pythagorean Theorem: a² + b² = c²

h² + 144² = 156²

h² + 20736 = 24336

h² = 3600

h = 60

x² + 60² = 65²

x² + 3600 = 4225

x² = 625

x = 25

3 0

3 years ago

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

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3 years ago

A committee of 3 is being formed randomly from the employees at a school: 5 administrators, 34 teachers, and 4 staff. What is th

pav-90 [236]

Answer:0.000810

Step-by-step explanation:

Given

A committee of 3 members is formed using

5 Admin, 34 teachers, and 4 staff

Favorable outcomes are selecting 3 members as Administrators

total members =5+34+4=43

Probability is given by

$\Rightarrow P=\dfrac{\text{Favorable}}{\text{Total}}$

$\Rightarrow P=\dfrac{^5C_3}{^{43}C_3}=\dfrac{10}{43\times 41\times 7}\\\\\Rightarrow P=\dfrac{10}{12,341}=0.000810$

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3 years ago

Question In Picture, Thanks.

Ainat [17]

6z to the 4th power - 2z to the 3rd power

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3 years ago

123567543 what is the value of 3

borishaifa [10]

Answer:

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Step-by-step explanation:

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3 years ago