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3 years ago

If Paul saves $3.00 every week for twelve weeks and then buys a CD for $15.00 plus 9% tax, how much money will he have left?

Mathematics

2 answers:

pochemuha3 years ago

8 0

Answer:

18.90

Step-by-step explanation:

3x12=36

36-15=21

21 x 0.9

ch4aika [34]3 years ago

8 0

Answer:

$19.65

Step-by-step explanation:

12*3=36

15*.09=1.35

15+1.35=16.35

36-16.35=$19.65

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PartA)What are the transformations are needed in in order to obtain the graph of G(x) from the graph of f(x) Select all that app

iVinArrow [24]

Part A)

Recall that:

1) The function represented by the graph of the function f(x) translated vertically n units up and horizontally m units left is:

$f(x+m)+n\text{.}$

2) The function represented by the graph of the function f(x) reflected over the x-axis is:

$-f(x)\text{.}$

Now, notice that g(x) is the function f(x) reflected over the x-axis and then translated vertically 6 units up and horizontally 4 units left.

Answer Part A:

Options B, C, and D.

Part B) To graph g(x) we will reflect the graph of f(x) over the x-axis and then we will translate it vertically 6 units up and horizontally 4 units left.

We know that the graph of f(x)=|x| is:

The above graph reflected over the x-axis is:

Finally, the above graph translated vertically 6 units up and horizontally 4 units left is:

Answer part B:

7 0

1 year ago

In the diagram below angle AED=(4n-20)° and angle BEC=(2n-16)°. What is the measure of angle CED in degrees

Pavlova-9 [17]

Check the picture below.

so, bearing in mind that vertical angles are equal, thus ∡AED = ∡BEC, and we know that ∡BEC = 52 = ∡AED, so those two angles added up are 104°.

now, a circle has a total of 360°, if we take away those 104° from the red angles, what's leftover is just 256°, well, the green angles there are also vertical angles and thus twins, so each one will take half of 256°.

∡BEA = 128° = ∡CED.

3 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: The probability of no arrivals in a one-minute period is 0.000045.

Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c: The probability of no arrivals in a 15-second is 0.0821.

Part d: The probability of at least one arrival in a 15-second period is 0.9179.

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

The probability of no arrivals in a one-minute period is 0.000045.

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

The probability of three or fewer passengers arrive in a one-minute period is 0.0103.

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

The probability of no arrivals in a 15-second is 0.0821.

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

The probability of at least one arrival in a 15-second period is 0.9179.

7 0

3 years ago

What is the solution to the equation 13.7y = 6.2y + 30?

il63 [147K]

Here, we are required to determine the solution to the equation: 13.7y = 6.2y + 30?

The solution to the equation is y = 7.5

The solution is thus,

13.7y = 6.2y + 30
13.7y - 6.2y = 30
7.5y = 30

Therefore, y = 30/7.5 = 4

Therefore, y = 4 is the solution to the equation.

Read more:

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8 0

2 years ago

How do i solve this kind of problem

Advocard [28]

Answer:

y=x-2

Step-by-step explanation:

to get -2 to -4 you would have to add -2

if we follow the pattern of subtracting 2 then you would see that -2 is b sense we are subtracting.

The m value doesn't have to change so we would leave the x how it is

The goal for this question is to find the pattern the numbers share though out the table.

4 0

3 years ago