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Kryger [21]
3 years ago
12

Solve these

}^{2} - 5x + 6" alt=" {x}^{2} - 5x + 6" align="absmiddle" class="latex-formula">
{x}^{2}  - 11x + 28
​
Mathematics
2 answers:
AlekseyPX3 years ago
7 0

Answer:

hope this will help you more

beks73 [17]3 years ago
7 0

Answer:

x^2-5x+6

x = 3, x = 2

x^2-11x+28

x = 7, x = 4

Step-by-step explanation:

One is asked to solve two quadratic equations. Both of these quadratic equations are in standard form. This means that the two equations follow the following general format;

ax^2+bx+c

The quadratic formula is a method of solving a quadratic equation using the coefficients of the terms in the equation. The quadratic equation is the following,

\frac{-b(+-)\sqrt{b^2-4ac}}{2a}

Substitute in each of the terms and solve for the roots in each equation;

x^2-5x+6

Substitute, and solve

\frac{-(-5)(+-)\sqrt{(-5)^2-4(1)(6)}}{2(1)}\\\\=\frac{5(+-)\sqrt{25-25}}{2}

=\frac{5+1}{2}          =\frac{5-1}{2}

=3               =2

x^2-11x+28

Substitute, and solve

\frac{-(-11)(+-)\sqrt{(-11)^2-4(1)(28)}}{2(1)}\\=\frac{11(+-)\sqrt{121-112}}{2}

=\frac{11+\sqrt{9}}{2}           =\frac{11-\sqrt{9}}{2}

=\frac{11+3}{2}              =\frac{11-3}{2}

=7                     =4

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Police response time to an emergency call is the difference between the time the call is first received by the dispatcher and th
Marizza181 [45]

Answer:

\mu = 8.8\\\sigma = 1.9

Now we are supposed to find probabilities that the response time is between 5 and 10 minutes i.e P(5<x<10)

Formula : z= \frac{x-\mu}{\sigma}

at x = 5

z= \frac{5-8.8}{1.9}

z=-2

at x = 10

z= \frac{10-8.8}{1.9}

z=0.6315

P(-2<z<0.6315)=P(z<0.6315)-P(z<-2)

Refer the z table

P(-2<z<10)=0.7357-0.0228=0.7129

So, the probability that response time is between 5 and 10 minutes  is 0.7129

b)the response time is less than 5 minutes

at x = 5

z= \frac{5-8.8}{1.9}

z=-2

P(x<5)=P(z<-2)=0.0228

So, the probability that  the response time is less than 5 minutes is 0.0228

c)the response time is more than 10 minutes

at x = 10

z= \frac{10-8.8}{1.9}

z=0.6315

P(x>10) = 1-P(x<10) = 1-P(z<0.63) = 1-0.7357 = 0.2643

So, The probability that  the response time is more than 10 minutes is 0.2643

4 0
3 years ago
Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).
Dvinal [7]
<span>For given hyperbola:
center: (0,0)
a=7 (distance from center to vertices)
a^2=49
c=9 (distance from center to vertices)
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Equation of given hyperbola:


..
2: vertices (0,+/-3) foci (0,+/-6)
hyperbola has a vertical transverse axis
Its standard form of equation:  , (h,k)=(x,y) coordinates of center
For given hyperbola:
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a=3 (distance from center to vertices)
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Equation of given hyperbola:
</span>
4 0
3 years ago
Mr. Brown owned a house, which he rented for $60 a month. The house was assessed at $9000. In 1975 the rate of taxation was incr
sergiy2304 [10]

Answer:

Raised by $2.25/month

Step-by-step explanation:

9000/1000=9

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7 0
3 years ago
Use the zero product property to find the solutions to the equation x2 – 13x + 30 = 0.
alukav5142 [94]

Answer:

<u>The solutions to this quadratic equation are 10 and 3.</u>

Step-by-step explanation:

1. Let's recall the formula for solving this type of equations, called quadratic equations:

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The equation given is  x2 – 13x + 30 = 0,

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Now replacing with the real values, we have

x = [ - (-13) +/- √ (-13)² - 4 * 1 * 30] / 2 * 1

x = [ 13 +/- √ 169 - 120] / 2

x = [13 +/- √ 49]/ 2

x = [ 13 +/- 7 ]/ 2

<u>x₁ = 13 + 7/2 = 20/2 = 10</u>

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valentinak56 [21]
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