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3 years ago

Can you guys help me with this it’s due in 5 min I need a correct answer.

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

4 0

Answer:

The answer is R=2

Step-by-step explanation:

30=2 times 15

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This week there are 144 members at the meeting, which is 90% of the total group. What is the total number of green team members

irina1246 [14]

Answer:

The total number of Green team members is 160.

Step-by-step explanation:

Given 90% of the total members present at the meeting which is 144.

Let the total number of members be x.

$90\%\ of x=144\\\frac{90\times\ x}{100}=144\\9x=1440\\x=\frac{1440}{9}=160$

Hence we can say that, The total number of Green team members is 160.

6 0

3 years ago

Find the area of the shaded region.<br> 8 in<br> 24 in

zloy xaker [14]

Answer:

96 in^2

Step-by-step explanation:

area of rectangle: 24*8=192

Area of triangle: (1/2)*base*height

= .5 * 8 * 24

=192/2

=96

area of shaded=rectangle - triangle

= 192-96

which is also just 192/2

so it is 96

4 0

3 years ago

Milo drives 126 miles on 7.5 gallons of gas. How far does he travel on each gallon of gas? 1.68 miles 16.8 miles 168 miles 1,680

Tanzania [10]

The answer is B. 16.8

8 0

3 years ago

Read 2 more answers

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

What number is 53% of 470?

Ivahew [28]

53% of 470 is exactly 249.10000000000002. What ever is closest to that is your answer.

3 0

3 years ago