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BartSMP [9]
2 years ago
13

The law of cosines is a^2+b^2-2abcosC=c^2. Find the value of 2abcosC.

Mathematics
2 answers:
Ghella [55]2 years ago
5 0

Answer:

(D)7=2abcosC

Step-by-step explanation:

It is given from figure that c=1, a=2 and b=2, thus using the law of cosines, we get

c^2=a^2+b^2-2abcosC

Substituting the given values, we get

(1)^2=(2)^2+(2)^2-2abcosC

1=4+4-2abcosC

1=8-2abcosC

1-8=-2abcosC

-7=-2abcosc

7=2abcosC

Thus, the value of 2abcosC is 7.

Hence, option (D) is correct.

polet [3.4K]2 years ago
4 0
2ab x cos (C) = 7 because

a^2 + b^2 - 2ab x cos(C) = c^2
2ab x cos(C) = a^2 + b^2 - c^2
2ab x cos(C) = 2^2 + 2^2 - 1^2
2ab x cos (C) = 7
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Step-by-step explanation:

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1 year ago
Researchers have a sample size of 24 , and they are using the Student's t-distribution. What are the degrees of freedom and how
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Answer:

For this case assuming that the random variable is X

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And replacing n = 24 we got:

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And we notate the distribution we got: X \sim t_{n-1}= t_{23}

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

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Solution to the problem

For this case assuming that the random variable is X

df = n-1

And replacing n = 24 we got:

df = 24-1=23

And we notate the distribution we got: X \sim t_{n-1}= t_{23}

7 0
2 years ago
All I need is this answer,please help!!
il63 [147K]
Hello there,

There is a total of 40 animals in that in 10 zebras, 10 lions, 10 elephants, and 10 monkeys.
So the probability of them both being lions is pretty rare. 2/40 so that would give you a 1/20 chance of both crackers being both lions.

I hope this helps!
7 0
2 years ago
A principal of $2000 is placed in a savings account at 3% per annum compounded annually. How much is in the account after one ye
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Answer:

Step-by-step explanation:

After one year

A=p(1+r/n)^nt

=2000(1+0.03/12)^12*1

=2000(1+0.0025)^12

=2000(1.0025)^12

=2000(1.0304)

=$2060.8

After two-years

A=p(1+r/n)^nt

=2060.8(1+0.03/12)^12*2

=2060.8(1+0.0025)^24

=2060.8(1.0025)^24

=2060.8(1.0618)

=$2188.157

After three years

A=p(1+r/n)^nt

=2188.157(1+0.03/12)^12*3

=2188.157(1+0.0025)^36

=2188.157(1.0025)^36

=2188.157(1.0941)

=$2394.063

8 0
2 years ago
A pet store has 10 ​puppies, including 3 ​poodles, 4 ​terriers, and 3 retrievers. if rebecka and​ aaron, in that​ order, each se
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P(both poodle) = (3/10)(3/10) = 9/100

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