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EleoNora [17]
2 years ago
5

I need help on this how do i do this

Mathematics
1 answer:
kap26 [50]2 years ago
7 0
Answer: 4311/9900 = 0.4354 4936/990 = 4.985 530/900 = 0.58 674/990 = 0.680 8099/990 = 8.180 explanation: divide the numbers and select the answers.
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My sister is half my age when I was 6 now im 70 how old is she?​
TEA [102]

Answer:

your sister would be 35

Step-by-step explanation:

5 0
2 years ago
Add and simplify 9/16+1/2 =
Ad libitum [116K]
Sorry for the bad handwriting.

3 0
3 years ago
How many digits are in 385,604 ?
Kaylis [27]
6 llllllllllllllllllll
6 0
2 years ago
Read 2 more answers
[Geometry Basics] I'm just checking if this is correct:
Leno4ka [110]
\bf ~~~~~~~~~~~~\textit{middle point of 2 points }
\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ x &,& y~) 
%  (c,d)
&&(~ -9 &,& -1~)
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{ x_2 +  x_1}{2}\quad ,\quad \cfrac{ y_2 +  y_1}{2} \right)
\\\\\\
\left( \cfrac{-9+x}{2}~~,~~\cfrac{-1+y}{2} \right)=\stackrel{midpoint}{(8,14)}\implies 
\begin{cases}
\cfrac{-9+x}{2}=8\\\\
-9+x=16\\
\boxed{x=25}\\
-------\\
\cfrac{-1+y}{2} =14\\\\
-1+y=28\\
\boxed{y=29}
\end{cases}

\bf -------------------------------\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
%  (a,b)
&&(~ x &,& y~) 
%  (c,d)
&&(~10 &,& 12~)
\end{array}\qquad
\\\\\\
\left( \cfrac{10+x}{2}~~,~~\cfrac{12+y}{2} \right)=\stackrel{midpoint}{(6,9)}\implies 
\begin{cases}
\cfrac{10+x}{2}=6\\\\
10+x=12\\
\boxed{x=2}\\
-------\\
\cfrac{12+y}{2} =9\\\\
12+y=18\\
\boxed{y=6}
\end{cases}
3 0
3 years ago
Let X denote a random variable which corresponds to the number of players in a team who get injured during a game. If we assume
Feliz [49]

Answer:

(a)

The values of X can be 0, 1, 2 , ..., 10 . So, X is a discrete random variable.

(b)

The distribution of X is Binomial distribution with the parameters n = 10 and p = 0.2

(c)

Probability that no one or one person will be injured = P(X = 0) + P(X = 1)

= 10C0 * 0.20 * (1 - 0.2)10-0 + 10C1 * 0.21 * (1 - 0.2)10-1

= 0.810 + 10 * 0.2 * 0.89

= 0.3758096

(d)

Average value of X = np

Average value of X = 10 * 0.2 = 2

(e)

Variance of X = np(1-p)

Variance of X = 10 * 0.2 * (1 - 0.2) = 1.6

(f)

Number of ways in which 2 people gets injured = 10C2 = 10! / ((10-2)! 2!) = (10 * 9) / (2 * 1) = 45

Assume the best player got injures, number of ways in which one people out of remaining 9 people gets injured = 9C1

= 9! / ((9-1)! 1!)

= 9

Probability that the best player got injured = Number of ways in which 1 people gets out of 9 and best person gets injured / Number of ways in which 2 people gets injured

= 9 / 45

= 0.2

4 0
3 years ago
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