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2 years ago

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Jenny is trimming the edge of pillows with lace. Each pillow requires 15 inches of lace. She has one piece of lace that is 3 fee

t long and another that is 4 feet long. What is the maximum number of pillows she will be able to trim with lace?
this is for Achieve3000

1 answer:

Anna71 [15]2 years ago

4 0

Answer:

I think Jenny will be able to do 9 pillows with the lace trim.

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Step-by-step explanation:

g(x) has clearly in its core the same function curve.

but it is

1. upside-down

2. moved up from the x-axis by 1 unit

3. moved to the right of the y-axis by 3 units

so, how do we express these 3 attributes in the functional definition ?

1. easily : by flipping the sign. g(x) = -x² is the same function type just upside-down (mirrored into the negative y space).

g(x) = -x²

2. also very easy : a function is moved up or down on the coordinate grid by adding (or subtracting) a constant.

we need to move our function up by 1 unit : we add 1.

that makes currently g(x) = -x² + 1

3. this is the trickiest part. to move a function left or right we need to make the function "think" that the input value x is not x, but it is (x ± constant).

let's ignore 1. and 2. for the moment and just focus moving the original function 3 units to the right.

that tells us that the functional result value of x in the shifted function must be the same as the functional result value in the original function for an input value that is 3 units "earlier" on the x-axis.

that would mean g(0) = f(-3), g(1) = f(-2), g(2) = f(-1), g(3) = f(0), ...

so, we see, g(x) = f(x-3) = (x-3)²

now, we combine again 1., 2. and 3., and we get

g(x) = -f(x-3) + 1 = -(x-3)² + 1

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2 months ago, Blake used his credit card for a transaction of 400 dollars. The bank charges a rate of interest of

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$206.27 per month is needed to payoff the balance in 2 months. Total interest is $12.54.

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

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Answer:

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

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

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

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

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,

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

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

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,

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

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

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:

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

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:

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

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

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

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

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