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4 years ago

8

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

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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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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4 years ago

Read 2 more answers

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

Which fraction has a value less than 5-7?

lesya [120]

Answer:

3/4

Step-by-step explanation:

4 0

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