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Rasek [7]
2 years ago
7

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<br> 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

<u>15x+6y=57</u><u> </u> +

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 &amp;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
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