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uranmaximum [27]

3 years ago

7

Solve x/-4 - (-8) =12 16 80 -80 -16

1 answer:

Pachacha [2.7K]3 years ago

7 0

Answer:

x = -16.

Step-by-step explanation:

x/-4 - (-8) = 12

x/-4 + 8 = 12

x/-4 = 12 - 8

x/-4 = 4

Cross multiply:

x = -4 * 4

x = -16.

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Find the solution to the equation below.<br> 2x^2+3x-20=0

mariarad [96]

Answer:

x = -4, 5/2

Step-by-step explanation:

A quadratic can be solved may ways, including graphing, factoring, and the quadratic formula. You can also check possible answers by making use of the relationships between solutions and the coefficients.

__

A graph is attached. It shows the solutions to be -4 and 5/2.

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When factored, the equation becomes ...

(2x -5)(x +4) = 0 . . . . . has solutions x=-4, x=5/2 (these make the factors zero)

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Using the quadratic formula, the solutions of ax^2 +bx +c = 0 are found from ...

x = (-b±√(b²-4ac))/(2a)

x = (-3±√(3²-4(2)(-20))/(2(2)) = (-3±√169)/4 = {-16, +10}/4

x = {-4, 5/2}

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For ax^2 +bx +c = 0, the solutions must satisfy ...

product of solutions is c/a = -20/2 = -10

Only the first and last choices have this product.

sum of solutions is -b/a = -3/2

Only the first choice (-4, 5/2) has this sum.

8 0

3 years ago

Suppose a random variable x is best described by a uniform probability distribution with range 22 to 55. Find the value of a tha

const2013 [10]

Answer:

(a) The value of <em>a</em> is 53.35.

(b) The value of <em>a</em> is 38.17.

(c) The value of <em>a</em> is 26.95.

(d) The value of <em>a</em> is 25.63.

(e) The value of <em>a</em> is 12.06.

Step-by-step explanation:

The probability density function of <em>X</em> is:

$f_{X}(x)=\frac{1}{55-22}=\frac{1}{33}$

Here, 22 < X < 55.

(a)

Compute the value of <em>a</em> as follows:

$P(X\leq a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.95\times 33=[x]^{a}_{22}\\\\31.35=a-22\\\\a=31.35+22\\\\a=53.35$

Thus, the value of <em>a</em> is 53.35.

(b)

Compute the value of <em>a</em> as follows:

$P(X< a)=\int\limits^{a}_{22} {\frac{1}{33}} \, dx \\\\0.95=\frac{1}{33}\cdot \int\limits^{a}_{22} {1} \, dx \\\\0.49\times 33=[x]^{a}_{22}\\\\16.17=a-22\\\\a=16.17+22\\\\a=38.17$

Thus, the value of <em>a</em> is 38.17.

(c)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.85=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.85\times 33=[x]^{55}_{a}\\\\28.05=55-a\\\\a=55-28.05\\\\a=26.95$

Thus, the value of <em>a</em> is 26.95.

(d)

Compute the value of <em>a</em> as follows:

$P(X\geq a)=\int\limits^{55}_{a} {\frac{1}{33}} \, dx \\\\0.89=\frac{1}{33}\cdot \int\limits^{55}_{a} {1} \, dx \\\\0.89\times 33=[x]^{55}_{a}\\\\29.37=55-a\\\\a=55-29.37\\\\a=25.63$

Thus, the value of <em>a</em> is 25.63.

(e)

Compute the value of <em>a</em> as follows:

$P(1.83\leq X\leq a)=\int\limits^{a}_{1.83} {\frac{1}{33}} \, dx \\\\0.31=\frac{1}{33}\cdot \int\limits^{a}_{1.83} {1} \, dx \\\\0.31\times 33=[x]^{a}_{1.83}\\\\10.23=a-1.83\\\\a=10.23+1.83\\\\a=12.06$

Thus, the value of <em>a</em> is 12.06.

7 0

3 years ago

Is (0,0) a solution to this system? y ≥ x^2 + x - 4, y> x^2 + 2x + 1

Ipatiy [6.2K]

Y≥x²+x-4
y>x²+2x+1

Substitute (0,0):

0≥-4 true
0>1 false

(0,0) is not a solution of the system.

3 0

3 years ago

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Complete the following statement given: QXR=NYC a.QX= ? b. Y=?

AfilCa [17]

Answer:

Step-by-step explanation:

3 0

3 years ago

The cost of a ticket to the soccer game is 6$. There are y Number of people in a group that want to go to the game. Which of the

erastovalidia [21]

If each ticket cost six dollars and there is y amount of people it would be six for every person.
your answer would be (C)6y

8 0

3 years ago