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Eva8 [605]
2 years ago
6

Word Problem:

Mathematics
2 answers:
stiv31 [10]2 years ago
5 0
81910 is the total answer
jonny [76]2 years ago
5 0

Answer:

the answer is 135.8 plzz mark me the brainliest

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A little boy mass 10kg is placed at a height of 2cm above the ground,what is the potential energy of the boy with reference to t
marin [14]
Potential gravitational energy:

mgh

10kg • 10m/s^2 • 0.02 =

2 Joules
3 0
3 years ago
A psychologist wants to know if the difficulty of a task influences our estimate of how long we spend working at it. She designs
Delicious77 [7]

Solution :

It is given that a psychologist wants to determine the difficulty of the task influences on the estimate of how long people spend working on it.

Now we have to describe a experiment by using a completely randomized design and learn the effect of the difficulty on the estimated time is by randomly assigning 15 students to a Group 1 of easy mazes and other 15 students to the second group 2 of hard mazes. And then we compare the time estimates of the two groups.

7 0
3 years ago
Which one is equivalent what’s the work that goes with it? i’m very confused
Elenna [48]

Answer:

0.8r

Step-by-step explanation:

r - 0.2r

= 1r - 0.2r ← factor out r from each term

= r(1 - 0.2)

= r × 0.8

= 0.8r

5 0
2 years ago
Use the following experiment. A state lottery game consists of choosing one card from each of the four suits in a standard deck
Masteriza [31]

Jupiter Saturn Mars Earth Sun

7 0
2 years ago
What is the formula for the expected number of successes in a binomial experiment with n trials and probability of success​ p? C
charle [14.2K]

Answer:

(D)E[ X ] =np.

Step-by-step explanation:

Given a binomial experiment with n trials and probability of success​ p,

f(x)=\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}, 0\leq  x\leq n

E(X)=\sum_{x=0}^{n}xf(x)= \sum_{x=0}^{n}x\left(\begin{array}{c}n\\k\end{array}\right)p^x(1-p)^{n-x}

Since each term of the summation is multiplied by x, the value of the term corresponding to x = 0 will be 0. Therefore the expected value becomes:

E(X)=\sum_{x=1}^{n}x\left(\begin{array}{c}n\\x\end{array}\right)p^x(1-p)^{n-x}

Now,

x\left(\begin{array}{c}n\\x\end{array}\right)= \frac{xn!}{x!(n-x)!}=\frac{n!}{(x-)!(n-x)!}=\frac{n(n-1)!}{(x-1)!((n-1)-(x-1))!}=n\left(\begin{array}{c}n-1\\x-1\end{array}\right)

Substituting,

E(X)=\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^x(1-p)^{n-x}

Factoring out the n and one p from the above expression:

E(X)=np\sum_{x=1}^{n}n\left(\begin{array}{c}n-1\\x-1\end{array}\right)p^{x-1}(1-p)^{(n-1)-(x-1)}

Representing k=x-1 in the above gives us:

E(X)=np\sum_{k=0}^{n}n\left(\begin{array}{c}n-1\\k\end{array}\right)p^{k}(1-p)^{(n-1)-k}

This can then be written by the Binomial Formula as:

E[ X ] = (np) (p +(1 - p))^{n -1 }= np.

5 0
3 years ago
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