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3 years ago

Babylon

Mathematics

2 answers:

masha68 [24]3 years ago

4 0

I need to see the table.

garri49 [273]3 years ago

3 0

We need the picture of the table to answer the question. If you can’t. I would just guess B is the answer.

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Can you please help? The question is in the picture picture. thank you

Svetllana [295]

Answer:

AB should be equal to 37

Step-by-step explanation:

3 0

3 years ago

Help if u know it please!!!

Ulleksa [173]

Mean: add all the numbers, then dived the sum by hwo many numbers there are.
72.3 + 74.5 + 81.1 + 72.3 + 75.6 + 79 = 754.8
754.8 ÷ 6 = 75.8
Mean = 75.8°

Median: Put all the number in order from lowest to highest. The middle number is the median. If there are two numbers in the middle, add them then divide the sum by 2.
72.3, 72.3, 74.5, 75.6, 79, 81.1
74.5 + 75.6 = 150.1
150.1 ÷ 2 = 75.05 ~ 75
Median = 75°

5 0

3 years ago

25 points please help thanks

weeeeeb [17]

Answer:

Step-by-step explanation:

With a slope of 1 and a y intercept of -3

y = 1x - 3

y = 1(-2) - 3

y = -5

(-2, -5)

6 0

2 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

4 years ago

Find the missing angles

Masteriza [31]

Answer:

b = 34°

Step-by-step explanation:

b + 115° + 31° = 180°

b + 146° = 180

b = 180° - 146°

b = 34°

3 0

3 years ago

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