I got stuck on this question - anyone willing to help?

The number of a checking account is also known as a.account number b.checking number c. Bank number d.routing number

Sophie [7]

Answer:

A account number

Step-by-step explanation:

8 0

3 years ago

A.How far is the library from the park?<br> B.How far is the park from the football field?

Scorpion4ik [409]

Answer:

a. The distance from the library to the park is 2√5 miles

b. The distance from the park to the football field is 2√35 miles

The last answer ⇒ 2√5 miles , 2√35 miles

Step-by-step explanation:

* Lets revise the rules in the right angle triangle when we draw the

perpendicular from the right angle to the hypotenuse

- In triangle ABC

# Angle B is a right angle

# The hypotenuse is AC

# BD ⊥ AC

∴ (AB)² = AD × AC

∴ (BC)² = CD × AC

∴ (BD)² = AD × CD

∴ BD × AC = AB × BC

* Lets use one of these rules to solve the problem

- Vertex B represents the position of the park

- Vertex A represents the position of the home

- Vertex C represents the position of the football field

- Vertex D represents the position of the library

- AB represents the distance from the home to the park

- CB represents the distance from the football field to the park

- DB represents the distance from the library to the park

- AC represents the distance from the home to the football field

- AD represents the distance from the home to the library

- CD represents the distance from the football field to the library

* Now lets solve the problem

a.

∵ The distance from the home to the library is 4 miles

∴ AD = 4

∵ The distance from the football field to the library is 10 miles

∴ CD = 10

∵ The distance from the library to the park represented by BD

- Lets use the rule (BD)² = AD × CD

∴ (BD)² = 4 × 10 = 40 ⇒ take √ for both sides

∴ BD = √40 = 2√5 miles

* The distance from the library to the park is 2√5 miles

b.

∵ The distance from the park to the football field represented by BC

* Lets use the rule (BC)² = CD × AC

∵ CD = 10

∵ AC = 4 + 10 = 14

∴ (BC)² = 10 × 14 = 140 ⇒ take √ for both sides

∴ BC = √140 = 2√35 miles

* The distance from the park to the football field is 2√35 miles

5 0

3 years ago

What is the value of x enter your awnser in the box<br>

Llana [10]

Answer:

9

Step-by-step explanation:ratios

7 0

3 years ago

Write the equation of a line with a slope of 5/2 and passes through the point (2,3)

tamaranim1 [39]

Answer:

y - 3 = ⁵/₂(x - 2) {the point-slope form of the equation}

y = ⁵/₂x - 2 {the slope-intercept form of the equation}

5x - 2y = 4 {the standard form of the equation}

Step-by-step explanation:

The point-slope form of equation is: y - y₀ = m(x - x₀), where (x₀, y₀) is any point the line passes through and m is the slope.

m = ⁵/₂

(2, 3) ⇒ x₀ = 2, y₀ = 3

So, the point-slope form of the equation:

y - 3 = ⁵/₂(x - 2)

Therefore:

y - 3 = ⁵/₂x - 5 {add 3 to both sides}

y = ⁵/₂x - 2 ← the slope-intercept form of the equation

-⁵/₂x + y = - 2 {multiply both sides by (-2)}

5x - 2y = 4 ← the standard form of the equation

5 0

3 years ago

The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.

Ulleksa [173]

Answer:

0.52763 is the probability that the time between the next two calls will be 54 seconds or less.

0.19285 is the probability that the time between the next two calls will be greater than 118.5 seconds.

Step-by-step explanation:

We are given the following information in the question:

The time between telephone calls to a cable television service call center follows an exponential distribution with a mean of 1.2 minutes.

The distribution function can be written as:

$f(x) = \lambda e^{-\lambda x}\\\text{where lambda is the parameter}\\\\\text{Mean} = \mu = \displaystyle\frac{1}{\lambda}\\\\\Rightarrow 1.2 = \frac{1}{\lambda}\\\\\lambda = 0.84 \\f(x) = 0.84 e^{0.84 x}$

The probability for exponential distribution is given as:

$P( x \leq a) = 1 - e^{\frac{-a}{\mu}}\\\\P(a \leq x \leq b) = e^{\frac{-a}{\mu} -\frac{-b}{\mu}}$

a) P( time between the next two calls will be 54 seconds or less)

$P( x \leq 0.9)\\= 1 - e^{\frac{\frac{-54}{60}}{1.2}} = 0.52763$

0.52763 is the probability that the time between the next two calls will be 54 seconds or less.

b) P(time between the next two calls will be greater than 118.5 seconds)

$p( x > \frac{118.5}{60}) = P(x > 1.975)\\\\ = 1 - P(x \leq 1.975) \\\\= 1 -1+ e^{\frac{-1.975}{1.2}}\\\\= 0.19285$

0.19285 is the probability that the time between the next two calls will be greater than 118.5 seconds.

6 0

3 years ago