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Y_Kistochka [10]

3 years ago

11

How do i solve this problem? ursula mixed in 3 1/8 cups of dry ingredients with 1 2/5 cups of liquids ingredients.

1 answer:

Yuri [45]3 years ago

4 0

Now she has a mixture with 4 21/40 cups of ingredients.

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Which of these phrases contain a variable? Check all that apply.

kaheart [24]

Answer:

The length of the hall way

the weight of the wombat

4 0

3 years ago

Read 2 more answers

25 POINTS SOMEONE HELP ME OUT WITH THIS PLEASEEEE BRAINLIEST ANSWER

zloy xaker [14]

1. $y=7x-1$

2. $y=3x+b$
<u> </u><u>Sub (1, -5) into equation to find '</u><em><u>b'</u></em><u />
$-5=3(1)+b$
$-8=b$

$y=3x-8$

3. $x=-7$

4. $y=4$

3 0

3 years ago

Find the coordinates of the point 7/10 of the way from A to B. a=(-3,-6) b=(12,4)

Artemon [7]

Answer:

The coordinates of $M$ are $x = \frac{15}{2}$ and $y = 1$ .

Step-by-step explanation:

Let be $A = (-3,-6)$ and $B = (12, 4)$ endpoints of segment AB and $M$ a point located 7/10 the way from A to B. Vectorially, we get this formula:

$\overrightarrow {AM} = \frac{7}{10}\cdot \overrightarrow {AB}$

$\vec M - \vec A = \frac{7}{10}\cdot (\vec B - \vec A)$

By Linear Algebra we get the location of $M$ :

$\vec M = \vec A + \frac{7}{10}\cdot (\vec B - \vec A)$

$\vec M = \vec A +\frac{7}{10}\cdot \vec B - \frac{7}{10}\cdot \vec A$

$\vec M = \frac{3}{10}\cdot \vec A + \frac{7}{10}\cdot \vec B$

If we know that $\vec A = (-3,-6)$ and $\vec B = (12, 4)$ , then:

$\vec M = \frac{3}{10}\cdot (-3,-6)+\frac{7}{10}\cdot (12,4)$

$\vec M = \left(-\frac{9}{10},-\frac{9}{5} \right)+\left(\frac{42}{5} ,\frac{14}{5} \right)$

$\vec M =\left(-\frac{9}{10}+\frac{42}{5} ,-\frac{9}{5}+\frac{14}{5} \right)$

$\vec M = \left(\frac{15}{2} ,1\right)$

The coordinates of $M$ are $x = \frac{15}{2}$ and $y = 1$ .

6 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

Explain how bias in the survey might have affected the results.

vladimir1956 [14]

B. because "boring" implies that all classical music is boring.

3 0

3 years ago