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2 years ago

What is 7584 divided by 3

Mathematics

2 answers:

Pachacha [2.7K]2 years ago

6 0

Answer:

2528

Step-by-step explanation:

well three goes into 7 twice carry the one, three goes into 15, 5 times, three goes into 8 twice carry the 2, and then 3 goes into 24 eight times

ELEN [110]2 years ago

6 0

Answer:

2528

Step-by-step explanation:

3 can go into 7584, 2528 times

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Select the postulate that is illustrated for the real numbers.

san4es73 [151]

Answer:

The commutative postulate for multiplication

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A certain 5-digit combination lock can be reset with a new combination. The lock uses numbers from 1 to 6. After a friend accide

daser333 [38]

The probability is (1/6)^5 power. Since the locks are independent of each other, you can multiply the probabilities together. There are 6 possible numbers for each lock and there are 5 locks. So the probability is (1/6)^5 power.<span />

5 0

3 years ago

I need alot of help in math

umka2103 [35]

Plug in the dimensions to the formula given using a handy dandy calculator for both problems.

1. The diameter of the bottom of the cone is 4 so the radius (half that) is 2. Volume = 1/3 * pi * (2)² * 6
= 1/3 * pi * 4 * 6
= 1/3 * pi * 24
= 8 * pi
≈ 25.1327 cm³

2. They give you all you need, the radius.
Volume = 4/3 * pi * (6)³
= 4/3 * pi * 216
= 288 * pi
≈ 904.7787 in³

3 0

3 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

here are 1515 people in an office with 55 different phone lines. If all the lines begin to ring at once, how many groups of 55 p

Mashcka [7]

Answer:

360,360 groups of 5 people.

Step-by-step explanation:

We have been given that there are 15 people in an office with 5 different phone lines. We are asked to find groups of 5 people that can answer these lines, if all the lines begin to ring at once.

We will use fundamental principle of counting to solve our given problem.

There are 15 people to answer 1st line, that will leave us with 14 people to answer 2nd line.

Now, we will have 13 people to answer 3rd line, that will leave us with 12 people to answer 4th line.

There are 11 people to answer 5th call.

So 5 lines can be answered in $15\times 14\times 13\times 12\times 11=360,360$ ways.

Therefore, 360,360 groups of 5 people can answer these lines.

6 0

3 years ago