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3 years ago

Chuck drank

Mathematics

1 answer:

Ludmilka [50]3 years ago

3 0

2 +3 + 7 + 5 = 17 which in mixed numbers is 4 integers 1/4

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(Please respond quickly!)

ki77a [65]

Answer:

$x=12$

Step-by-step explanation:

we know that

A linear pair is two angles that are adjacent and form a line. The angle measure of a line is 180 degrees

In this problem we have that

$(10x-20)^o+(6x+8)^o=180^o$ --- by supplementary angles (form a linear pair)

solve for x

$(16x-12)=180$

$16x=180+12$

$16x=192$

$x=12$

3 0

3 years ago

Use the two-way table below to create a relative frequency table. Remember to round to the nearest percent.

Tems11 [23]

I don’t understand?

7 0

3 years ago

Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One

Naily [24]

Answer:

The expected value of X is $E(X)=\frac{2754}{73} \approx 37.73$ and the variance of X is $Var(X)=\frac{226192}{5329} \approx 42.45$

The expected value of Y is $E(Y)=\frac{73}{2} \approx 36.5$ and the variance of Y is $Var(Y)=\frac{179}{4} \approx 44.75$

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and probability mass function p(x). The expected value, denoted by E(X) or $\mu_x$ , is

$E(X)=\sum_{x\in D} x\cdot p(x)$

The probability mass function $p_{X}(x)$ of X is given by

$p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}$

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function $p_{Y}(x)$ of Y is given by

$p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}$

The expected value of X is

$E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)$

$E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73$

The expected value of Y is

$E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)$

$E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5$

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

$V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2$

The variance of X is

$E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)$

$E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}$

$Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45$

The variance of Y is

$E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)$

$E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377$

$Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75$

8 0

3 years ago

PLEASE HELP

SVETLANKA909090 [29]

Answer:

Hill 1: F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 2: F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 3: F(x) = 4(x - 2)(x + 5)

Step-by-step explanation:

Hill 1

You must go up and down to make a peak, so your function must cross the x-axis six times. You need six zeros.

Also, the end behaviour must have F(x) ⟶ -∞ as x ⟶ -∞ and F(x) ⟶ -∞ as x⟶ ∞. You need a negative sign in front of the binomials.

One possibility is

F(x) = -(x + 4)(x + 3)(x + 1)(x - 1)(x - 3)(x - 4)

Hill 2

Multiplying the polynomial by -½ makes the slopes shallower. You must multiply by -2 to make them steeper. Of course, flipping the hills converts them into valleys.

Adding 3 to a function shifts it up three units. To shift it three units to the right, you must subtract 3 from each value of x.

The transformed function should be

F(x) = -2(x +1)(x)(x -2)(x -3)(x - 6)(x - 7)

Hill 3

To make a shallow parabola, you must divide it by a number. The factor should be ¼, not 4.

The zeroes of your picture run from -4 to +7.

One of the zeros of your parabola is +5 (2 less than 7).

Rather than put the other zero at ½, I would put it at (2 more than -4) to make the parabola cover the picture more evenly.

The function could be

F(x) = ¼(x - 2)(x + 5).

In the image below, Hill 1 is red, Hill 2 is blue, and Hill 3 is the shallow black parabola.

6 0

3 years ago

Use the drawing tool(s) to form the correct answer on the provided graph.

NikAS [45]

Answer:

Here's one way to do it

Step-by-step explanation:

1. Solve the inequality for y

5x - y > -3

-y > -5x - 3

y < 5x + 3

2. Plot a few points for the "y =" line

I chose

\begin{gathered}\begin{array}{rr}\mathbf{x} & \mathbf{y} \\-2 & -7 \\-1 & -2 \\0 & 3 \\1 & 8 \\2 & 13 \\\end{array}\end{gathered}

−2

−1

−7

−2

You should get a graph like Fig 1.

3. Draw a straight line through the points

Make it a dashed line because the inequality is "<", to show that points on the line do not satisfy the inequality.

See Fig. 2.

4. Test a point to see if it satisfies the inequality

I like to use the origin,(0,0), for easy calculating.

y < 5x + 3

0 < 0 + 3

0 < 3. TRUE.

The condition is TRUE.

Shade the side of the line that contains the point (the bottom side).

And you're done (See Fig. 3).

6 0

3 years ago