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SpyIntel [72]
2 years ago
14

Two teams are going on a trip to playa game. There will be

Mathematics
2 answers:
Sholpan [36]2 years ago
3 0

Answer:

100

Step-by-step explanation:

60*2=120

10 seats taken up on 2 buses=20

120-20=100

user100 [1]2 years ago
3 0

Answer:

the last one (100) 60+60 equals120 10+ 10 equals 20 120-20 equals 100

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Lois says any addition equation where the addends are all the same can be written as a multiplication equation. Is Lois correct
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Lois is correct. This is because multiplication is repeated addition. It means that when your addends are just the same number but added several times, the sum is just equivalent to the product of that number times the number of times it was added. For example, 4+4+4 = 4×3 = 12.
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What is the equation of this graphed line ?
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Equations of straight lines are in the form y = mx + c (m and c are numbers). m is the gradient of the line and c is the y-intercept (where the graph crosses the y-axis).
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2 years ago
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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
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Answer:

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What is the area of a rhombus with dimensions of 22 inches and 10 inches?
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The answer to the problem is 220
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