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2 years ago

Ian owns a small business selling used books. He knows that in the last week 19

Mathematics

1 answer:

Softa [21]2 years ago

5 0

Step-by-step explanation:

last week there were 19+30+77 = 126 customers.

19/126 = 1/14 paid cash.

30/126 = 5/21 paid by debit card.

77/126 = 11/18 paid by credit card.

next week he expects 1800 customers.

that is compared to 126 a factor of

1800/126 = 300/21 = 100/7

now we actually only need to apply the same factor to any part of the total sum to get their share in the new total sum.

it simply follows the rule

(a + b + c) × d = ad + bd + cd

so,

77 × 100/7 = 11 × 100 = 1100 customers are expected to pay per credit card.

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6 high school seniors choose from among 20 quotes for their yearbook. What is the probability that at least 2 of them choose the

shusha [124]

Using the binomial distribution, it is found that there is a 0.0328 = 3.28% probability that at least 2 of them choose the same quote.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

In this problem, we have that:

There are 6 students, hence n = 6.

There are 20 quotes, hence the probability of each being chosen is p = 1/20 = 0.05.

The probability of one quote being chosen at least two times is given by:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

P(X < 2) = P(X = 0) + P(X = 1).

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.05)^{0}.(0.95)^{6} = 0.7351$

$P(X = 1) = C_{6,1}.(0.05)^{1}.(0.95)^{5} = 0.2321$

Then:

P(X < 2) = P(X = 0) + P(X = 1) = 0.7351 + 0.2321 = 0.9672.

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.9672 = 0.0328$

0.0328 = 3.28% probability that at least 2 of them choose the same quote.

More can be learned about the binomial distribution at brainly.com/question/24863377

6 0

2 years ago

Express 80 miles per hour in kilometers per hour

Anna [14]

80 miles per hour is 128.74752 kilometers per hour

8 0

3 years ago

Read 2 more answers

Mark all the statements that are true <br> a) the domain for this function is the set {3}

telo118 [61]

Answer:

the answer is 6 statements

Step-by-step explanation:

8 0

2 years ago

Find the work done by F= (x^2+y)i + (y^2+x)j +(ze^z)k over the following path from (4,0,0) to (4,0,4)

babunello [35]

$\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k$

We want to find $f(x,y,z)$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=x^2+y$

$\dfrac{\partial f}{\partial y}=y^2+x$

$\dfrac{\partial f}{\partial z}=ze^z$

Integrating both sides of the latter equation with respect to $z$ tells us

$f(x,y,z)=e^z(z-1)+g(x,y)$

and differentiating with respect to $x$ gives

$x^2+y=\dfrac{\partial g}{\partial x}$

Integrating both sides with respect to $x$ gives

$g(x,y)=\dfrac{x^3}3+xy+h(y)$

Then

$f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)$

and differentiating both sides with respect to $y$ gives

$y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C$

So the scalar potential function is

$\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}$

By the fundamental theorem of calculus, the work done by $\vec F$ along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it $L$ ) in part (a) is

$\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}$

and $\vec F$ does the same amount of work over both of the other paths.

In part (b), I don't know what is meant by "df/dt for F"...

In part (c), you're asked to find the work over the 2 parts (call them $L_1$ and $L_2$ ) of the given path. Using the fundamental theorem makes this trivial: