What are the roots of the quadratic function f(b)=b^2-75

If bd = 7x-10, bc = 4x-29, and cd = 5x-9, find each value

Airida [17]

Answer:

$bd=88$ , $bc=27$ , $cd=61$

Step-by-step explanation:

In this problem we have that

$bd=bc+cd$

we have

$bd=7x-10$

$bc=4x-29$

$cd=5x-9$

substitute the values and solve for x

$7x-10=(4x-29)+(5x-9)$

Group terms that contain the same variable in the right sides

$7x-10=(4x+5x)+(-29-9)$

Combine like terms in the right side

$7x-10=9x-38$

$9x-7x=-10+38$

$2x=28$

$x=14$

Find the value of bd

$bd=7(14)-10=88$

Find the value of bc

$bc=4(14)-29=27$

Find the value of cd

$cd=5(14)-9=61$

4 0

2 years ago

The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour. Round the

navik [9.2K]

Answer:

a) 0.019

b) 0.563

c) x = 1.966 hours

Step-by-step explanation:

E(X) = 1

Exponential random variable's probability function is given as

P(X=x) = λ e^(-λ.x)

The cumulative distribution function is given as

P(X ≤ x) = 1 - e^(-λ.x)

a) The time between the arrivals of small aircraft at a county airport that is exponentially distributed.

But the number of planes that land every hour will be obtained using the Poisson distribution formula.

It is the best for discrete systems.

Poisson distribution formula is given as

P(X = x) = (e^-λ)(λˣ)/x!

λ = 1 aircraft per hour.

The probability that more than three aircraft arrive within an hour = P(X > 3)

P(X > 3) = 1 - P(X ≤ 3) = 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)]

P(X > 3) = 1 - 0.98101 = 0.01899 = 0.019 to 3 d.p

b) If 30 separate one-hour intervals are chosen, what Is the probability that no interval contains more than three arrivals

Probability of one 1-hour interval not containing more than 3 arrivals = 1 - P(X > 3)

= 1 - 0.01899 = 0.98101

Probability that thirty 1-hour intervals will not contain more than 3 arrivals = (0.98101)³⁰ = 0.5626 = 0.563 to 3 d.p

c) Determine the length of an interval of time (In hours) such that the probability that no arrivals occur during the interval is 0.14

We can now use the cumulative distribution function for exponential random variable for this

P(X ≤ x) = 1 - e^(-λ.x)

P(X > x) = 1 - P(X ≤ x)

P(X > x) = e^(-λ.x)

λ = 1, x = ?,

0.14 = e⁻ˣ

e⁻ˣ = 0.14

In e⁻ˣ = In 0.14 = -1.966

-x = -1.966

x = 1.966 hours

Hope this Helps!!!

4 0

2 years ago

Peter mixes 4 1/2 cups of orange juice, 2 1/4 of ginger ale and 6 1/3 cups of strawberry lemonade to make some punch. What is th

ruslelena [56]

To add these amounts together, we must first find their least common multiple in order to get common denominators (b/c when you add fractions, the denominators must be the same).

We'll start by listing some of their multiples.
To do this, count by whatever the denominator is:
4 1/2 (denominator is 2): 2 4 6 8 10 12 14
2 1/4 (denominator is 4): 4 8 12 16
6 1/3 (denominator is 3): 3 6 9 12 15

Look and see which is the first multiple that all three denominators have. Circle them if it helps you. In this case, it's 12.

So now we have to multiply the denominators by whatever number it takes to reach 12, and multiply by the same number to the numerator:

4 1/2 (times 6 to both top and bottom) =
4 6/12

2 1/4 (times 3) = 2 3/12

6 1/3 (times 4) = 6 4/12

Add all these fractions together, and you get 12 13/12, which is equal to 13 1/12.

Thus, Peter makes a total of 13 1/2 cups.

Hope this made sense! tell me if anything is confusing/incorrect :))

8 0

1 year ago

How does the area of triangle RST compare to the area of triangle LMN? The area of △RST is 2 square units less than the area of

sashaice [31]

Inscribe triangle RST in the square with dimensions 4×4, as shown in the figure.

from the area of this square, 4*4=16, we remove the triangles with dimensions
3×4, 2×1 and 2×4, whose side lengths are shown in the figure, and we are left with the area of triangle RST.

so $Area(RTS)=16- \frac{1}{2}*3*4- \frac{1}{2}*2*1- \frac{1}{2}*2*4=16-6-1-4=5$ units squared

similarly,

$Area(LMN)=4*3- \frac{1}{2}*1*4- \frac{1}{2}*2*2- \frac{1}{2}*2*3=12-2-2-3=5$ units squared

Thus, the areas are equal.

8 0

1 year ago

Read 2 more answers

Find the distance from (4, −7, 6) to each of the following.

LenKa [72]

Answer:

(a) 6 units

(b) 4 units

(c) 7 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

(a) The distance from (4, -7, 6) to the xy-plane

The xy-plane is the point where z is 0. i.e

xy-plane = (4, -7, 0).

Therefore, the distance d is from (4, -7, 6) to (4, -7, 0)

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Hence, the distance to the xy plane is 6 units

(b) The distance from (4, -7, 6) to the yz-plane

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from (4, -7, 6) to (0, -7, 6)

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

(c) The distance from (4, -7, 6) to the xz-plane

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from (4, -7, 6) to (4, 0, 6)

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Hence, the distance to the xz plane is 7 units

(d) The distance from (4, -7, 6) to the x axis

The x axis is the point where y and z are 0. i.e

x-axis = (4, 0, 0).

Therefore, the distance d is from (4, -7, 6) to (4, 0, 0)

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

d = √(85)

d = 9.22

Hence, the distance to the x axis is 9.22 units

(e) The distance from (4, -7, 6) to the y axis

The x axis is the point where x and z are 0. i.e

y-axis = (0, -7, 0).

Therefore, the distance d is from (4, -7, 6) to (0, -7, 0)

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Hence, the distance to the y axis is 7.21 units

(f) The distance from (4, -7, 6) to the z axis

The z axis is the point where x and y are 0. i.e

z-axis = (0, 0, 6).

Therefore, the distance d is from (4, -7, 6) to (0, 0 6)

d = √[(4 - 0)² + (-7 - (0))² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Hence, the distance to the z axis is 8.06 units

5 0

2 years ago