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Bad White [126]
3 years ago
6

Select the table of values where the quadratic function changes direction at a different value of x than the others. A. x -3 -2

-1 0 1 2 3 y 28 19 12 7 4 3 4 B. x -3 -2 -1 0 1 2 3 y 4 3 4 7 12 19 28 C. x -3 -2 -1 0 1 2 3 y 2 3 2 -1 -6 -13 -22 D. x -3 -2 -1 0 1 2 3 y 1 -1 1 7 17 31 49
Mathematics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

D. hope the best for you.

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So what you will need to do is subtract 2x - 3x = -x, then subtract three from both sides 8 - 3 and the answer will be -x + 5
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5.1x=-15.3 solve for x
LekaFEV [45]

Answer


x=-3


Hope it helps!!

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Find the area A of triangle JKL with the vertices J(5,8), K(0,7), and L(5,4)
eimsori [14]

Answer:

Area of the triangle is 10 square units

Step-by-step explanation:

Recall that the formula for the area of a triangle is: "base x height / 2", and  in this case, we can consider the triangle's base as the segment that joins the vertices (5, 4) and (5, 8) which gives a segment length of 4 units. The height of the triangle is then the distance between the third vertex (0,7) and the base, which is exactly 5 units. Then the area becomes:

Area = 4 x 5 / 2 = 10 square units.

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3 years ago
A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature
Westkost [7]

Answer:

(a) The value of <em>k</em> is \frac{1}{15}.

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d)  P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 is not the pmf of <em>y</em>.

Step-by-step explanation:

The random variable <em>Y</em> is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of <em>k</em>.

\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}

Thus, the value of <em>k</em> is \frac{1}{15}.

(b)

Compute the value of P (Y ≤ 3) as follows:

P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\   =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 to be the pmf of Y it has to satisfy the conditions:

  1. P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\
  2. \sum P(y)=1

<u>Check condition 1:</u>

y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0

Condition 1 is fulfilled.

<u>Check condition 2:</u>

\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1

Condition 2 is not satisfied.

Thus, P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5 is not the pmf of <em>y</em>.

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3 years ago
Solve for w in terms of v, x, and y.<br> y = -XWV<br> W =
Elena L [17]

Answer:

w=21

Step-by-step explanation:

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