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iren [92.7K]
4 years ago
8

What is the answer to this math problem 3/10+2/5

Mathematics
2 answers:
romanna [79]4 years ago
7 0
7/10
3/10+(2)2/5
3/10+4/10
7/10
Dominik [7]4 years ago
4 0
Multiply 2/5 by 2 to get 4/10

4/10 + 3/10 = 7/10
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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.

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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}

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Answer:

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Step-by-step explanation:

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3 years ago
An investigator compares the durability of two different compounds used in the manufacture of a certain automobile brake lining.
Tanya [424]

Answer:

98%  Confidencce Interval is ( 3030.6, 7467.4 )

Step-by-step explanation:

Given that:

Sample size n_1 = 71

Sample size n_2 = 31

Sample mean \overline x_1 = 41628

Sample mean x_2 = 36,379

Population standard deviation \sigma_1 = 4934

Population standard deviation \sigma_2 = 4180

At 98% confidence interval level, the level of significcance = 1 - 0.98 = 0.02

Critical value at z_{0.02/2} = 2.33

The Margin of Error = z \times \sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma^2_2}{n_2} }

= 2.33 \times \sqrt{\dfrac{4934^2}{71}+\dfrac{4180^2}{31} }

= 2.33 \times \sqrt{\dfrac{24344356}{71}+\dfrac{17472400}{31} }

= 2.33 \times \sqrt{906504.06 }

= 2218.40

The Lower limit = ( \overline x_1 - \overline x_2) - (Margin \ of \ error)

= ( 41628 - 36379 ) - ( 2218.40)

= 5249 - 2218.40

= 3030.6

The upper limit = ( \overline x_1 - \overline x_2) + (Margin \ of \ error)

= ( 41628 - 36379 ) + ( 2218.40)

= 5249 + 2218.40

= 7467.4

∴  98%  Confidencce Interval is ( 3030.6, 7467.4 )

5 0
3 years ago
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