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Natasha_Volkova [10]
3 years ago
10

30.00 for a dozen regular cupcake

Mathematics
1 answer:
Leona [35]3 years ago
8 0

Each  cupcake is $2.50.  Divide the amount by 12



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The diagram below shows the contents of a jar from which you select marbles at random.
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Shanley would like to give $5 gift cards
Serjik [45]

Answer:

$5x + $4y ≤ $120

let x = 10

y = 5

$70  ≤ 120

Step-by-step explanation:

Note that

> means greater than

< means less than

≥  means greater or equal to

≤ means less than or equal to

the total amount Shanley has is $120. She can spend less than this amount or the amount exactly. So, the inequality sign to be used is ≤

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4 0
3 years ago
Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join
zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx  = 1

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y)  and replacing a and b with their respective values, we have

\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx  = 1

Since c is a constant, we can bring it out of the integral sign; to give us

c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx  = 1

Open the bracket

c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx  = 1

Integrate with respect to y

c\int\limits^3_0 {\frac{xy^{2}}{2}  +\frac{xy^{3}}{3} } \, dx (0,3}) = 1

Substitute 0 and 3 for y

c\int\limits^3_0 {(\frac{x* 3^{2}}{2}  +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2}  +\frac{x * 0^{3}}{3})} \, dx = 1

c\int\limits^3_0 {(\frac{x* 9}{2}  +\frac{x * 27}{3} ) - (0  +0) \, dx = 1

c\int\limits^3_0 {(\frac{9x}{2}  +\frac{27x}{3} )  \, dx = 1

Add fraction

c\int\limits^3_0 {(\frac{27x + 54x}{6})  \, dx = 1

c\int\limits^3_0 {\frac{81x}{6}  \, dx = 1

Rewrite;

c\int\limits^3_0 (81x * \frac{1}{6})  \, dx = 1

The \frac{1}{6} is a constant, so it can be removed from the integral sign to give

c * \frac{1}{6}\int\limits^3_0 (81x )  \, dx = 1

\frac{c}{6}\int\limits^3_0 (81x )  \, dx = 1

Integrate with respect to x

\frac{c}{6} *  \frac{81x^{2}}{2}   (0,3)  = 1

Substitute 0 and 3 for x

\frac{c}{6} *  \frac{81 * 3^{2} - 81 * 0^{2}}{2}    = 1

\frac{c}{6} *  \frac{81 * 9 - 0}{2}    = 1

\frac{c}{6} *  \frac{729}{2}    = 1

\frac{729c}{12}    = 1

Multiply both sides by \frac{12}{729}

c    =  \frac{12}{729}

c    =  0.0165 (Approximately)

8 0
3 years ago
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