Hello,

Vertices are on a line parallele at ox (y=-3)

The hyperbola is horizontal.

Equation is (x-h)²/a²- (y-k)²/b²=1

Center =middle of the vertices=((-2+6)/2,-3)=(2,-3)
(h+a,k) = (6,-3)
(h-a,k)=(-2,-3)
==>k=-3 and 2h=4 ==>h=2
==>a=6-h=6-2=4 (semi-transverse axis)

Foci: (h+c,k) ,(h-c,k)
h=2 ==>c=8-2=6

c²=a²+b²==>b²=36-4²=20

Equation is:
$\boxed{ \dfrac{(x-2)^2}{16} - \dfrac{(y+3)^2}{20} =1}

$

8 0

3 years ago

Steve got his car fixed for $557. After taxes it cost him $594. What was the tax rate to the nearest %?

I am Lyosha [343]

You're nearly there. Just some more steps.
557 * x = 594 tells us
x = 594 / 557 (divide both sides by 557)
x ≈ 1.066427

When we've found this number we must realise the following:
the '1' in this answer is the 100% the car fixing costs (without taxes)
the '0.66427' stands for the tax rate.

To find the tax rate we multiply 0.66427 by 100, which gives us:
0.66427 * 100 = 6.6427 ≈ 7

So, rounded to the nearest %, the tax rate is 7%

6 0

3 years ago

Consider the continuous random variable x, which has a uniform distribution over the interval from 110 to 150. The probability t

Virty [35]

Answer:

The probability that x will take on a value between 120 and 125 is 0.14145

Step-by-step explanation:

For uniform distribution between a & b

Mean, xbar = (a + b)/2

Standard deviation, σ = √((b-a)²/12)

For 110 and 150,

Mean, xbar = (150 + 110)/2 = 130

Standard deviation, σ = √((150-110)²/12 = 11.55

To find the probability that x will take on a value between 120 and 125

We need to standardize 120 & 125

z = (x - xbar)/σ = (120 - 130)/11.55 = - 0.87

z = (x - xbar)/σ = (125 - 130)/11.55 = - 0.43

P(120 < x < 125) = P(-0.87 < x < -0.43)

We'll use data from the normal probability table for these probabilities

P(120 < x < 125) = P(-0.87 < x < -0.43) = P(z ≤ -0.43) - P(z ≤ -0.86) = 0.33360 - 0.19215 = 0.14145

Hope this Helps!!!

3 0

3 years ago