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2 years ago

In triangle ABC, if b is the longest side, what can you conclude about angle A?

Mathematics

2 answers:

Yuri [45]2 years ago

5 0

Answer:

It is less than 90degrees

Step-by-step explanation:

malfutka [58]2 years ago

5 0

It is less than 90 degrees

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PLease NEED HELP! thank you

adelina 88 [10]

.32 is the answer to this question

7 0

2 years ago

7x-6y=9 5x+2y=19 Solve the system using elimination.

ikadub [295]

Step-by-step explanation:

multiply 5x+2y=19 by 3 youll get 15x+6y=57

eliminate y

7x-6y=9

15x+6y=57 +

22x=66

x=3

substitute x to 5x+2y=19

15+2y=19

2y=4

y=2

4 0

2 years ago

Alec pours the same amount of soup into 3 bowls. He used 4 cups of soup. How much soup is in each bowl?

Mrac [35]

1 3/4 cups of soup per bowl

4 0

3 years ago

Read 2 more answers

the product of five consecutive odd integers is 945 what is the greatest possible value of any of these integers

Troyanec [42]

If 945 is the product of 5 CONSECUTIVE ODD numbers, let's find the prime factors of 945:

945 = 5 x 7 x 27 , but we need 5 odd numbers:

945= 5 x 7 x ( 3 x 9 ) we still need one factor. This factor cannot be but 1

945 = 1 x 3 x 5 x 7 x 9 = 945 and the greatest value of these integers is 9

4 0

2 years ago

Find the mean, variance &a standard deviation of the binomial distribution with the given values of n and p.

MrMuchimi

A random variable following a binomial distribution over $n$ trials with success probability $p$ has PMF

$f_X(x)=\dbinom nxp^x(1-p)^{n-x}$

Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.

$\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1$

The mean is given by the expected value of the distribution,

$\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}$
$\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}$

The remaining sum has a summand which is the PMF of yet another binomial distribution with $n-1$ trials and the same success probability, so the sum is 1 and you're left with

$\mathbb E(x)=np=126\times0.27=34.02$

You can similarly derive the variance by computing $\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$ , but I'll leave that as an exercise for you. You would find that $\mathbb V(X)=np(1-p)$ , so the variance here would be

$\mathbb V(X)=125\times0.27\times0.73=24.8346$

The standard deviation is just the square root of the variance, which is

$\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834$

7 0

3 years ago