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3 years ago

If Robert buys a candy bar for 78 cents, how much change does he get back from a one-dollar bill?*

Mathematics

2 answers:

allsm [11]3 years ago

4 0

He will get back 22 cents Hope this helps

KengaRu [80]3 years ago

3 0

Lets do something simple. We have $1.00. We will use the one dollar. $1.00 will be there and then we have 78 cents used. So then you would do $1.00-.78= That equals $0.22. I hope this helped :)

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Help me out with these please!<br><br> No links to suspicious sites!

weeeeeb [17]

Answer:

1) N = 2. 2) -7/3

Step-by-step explanation:

6 = 3n

6/3 = n

2 = n

n = 2

-3 = 3t + 4

3t + 4 = -3

3t = -3 - 4

3t = -7

T = -7/3

5 0

2 years ago

Read 2 more answers

There are 4 people at a party. Consider the random variable X=’number of people having the same birthday ’ (match only month, N=

yulyashka [42]

Answer:

S = {0,2,3,4}

P(X=0) = 0.573 , P(X=2) = 0.401 , P(x=3) = 0.025, P(X=4) = 0.001

Mean = 0.879

Standard Deviation = 1.033

Step-by-step explanation:

Let the number of people having same birth month be = x

The number of ways of distributing the birthdays of the 4 men = (12*12*12*12)

The number of ways of distributing their birthdays = 12⁴

The sample space, S = { 0,2,3,4} (since 1 person cannot share birthday with himself)

P(X = 0) = $\frac{12P4}{12^{4} }$

P(X=0) = 0.573

P(X=2) = P(2 months are common) P(1 month is common, 1 month is not common)

P(X=2) = $\frac{3C2 * 12P2}{12^{4} } + \frac{4C2 * 12P3}{12^{4} }$

P(X=2) = 0.401

P(X=3) = $\frac{4C3 * 12P2}{12^{4} }$

P(x=3) = 0.025

P(X=4) = $\frac{12}{12^{4} }$

P(X=4) = 0.001

Mean, $\mu = \sum xP(x)$

$\mu = (0*0.573) + (2*0.401) + (3*0.025) + (4*0.001)\\\mu = 0.879$

Standard deviation, $SD = \sqrt{\sum x^{2} P(x) - \mu^{2}} \\SD =\sqrt{ [ (0^{2} * 0.573) + (2^{2} * 0.401) + (3^{2} * 0.025) + (4^{2} * 0.001)] - 0.879^{2}}$