Ask question

Ask question

All categories

3 years ago

11

5. You are making your Halloween costume this year. You have 6 yards of fabric and the pattern says that you need to cut it in t

wo pieces where one is two feet longer than the other. How long is the shorter piece of fabric going to be?

1 answer:

REY [17]3 years ago

7 0

Answer:

The shorter side piece of fabric = x = 8 feet

The longer side piece of fabrics = (x + 2) = 10 feet

Step-by-step explanation:

Total yards of fabrics = 6 yards

you need to cut it in two pieces where one is two feet longer than the other.

First piece= x feet

Second piece= x+2 feet

Convert 6 yards to feet

6 yards = 18 feet

First piece + second piece = total piece of fabrics

x + (x+2) = 18

x + x + 2 = 18

2x + 2 = 18

2x = 18 - 2

2x = 16

Divide both sides by 2

x = 16 / 2

x = 8 feet

8 feet = 2.66667 yards

x + 2

= 8 + 2

= 10 feet

10 feet = 3.33333 yards

The shorter side piece of fabric = x = 8 feet

The longer side piece of fabrics = (x + 2) = 10 feet

You might be interested in

NEED HELP ASAP 20POINTS The energy (and cost) needed to operate an air conditioner is proportional to the volume of a space that

insens350 [35]

Answer:

The daily cost of cooling the large warehouse with the same model of air conditioner is $\$1,953$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

In this problem

The scale factor is equal to 2.5

so

The scale factor elevated to the cube is $2.5^{3} =15.625$

To estimate the daily cost of cooling the large warehouse with the same model of air conditioner, multiply $125.00 by the scale factor elevated to the cube

so

$\$125(15.625)=\$1,953$

4 0

3 years ago

An article suggests that a poisson process can be used to represent the occurrence of structural loads over time. suppose the me

kirill115 [55]

Answer:

a) $\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

b) $P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

c) $e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution"

Solution to the problem

Let X our random variable who represent the "occurrence of structural loads over time"

For this case we have the value for the mean given $\mu = 0.5$ and we can solve for the parameter $\lambda$ like this:

$\frac{1}{\lambda} = 0.5$

$\lambda =2$

So then $X(t) \sim Poi (\lambda t)$

X follows a Poisson process

Part a

For this case since we are interested in the number of loads in a 2 year period the new rate would be given by:

$\lambda_1 = 2*2 = 4$

And let X our random variable who represent the "occurrence of structural loads over time" we know that:

$X(2) \sim Poi (4)$

And the expected value is $E(X) = \lambda =4$

So we expect 4 number of loads in the 2 year period.

Part b

For this case we want the following probability:

$P(X(2) >6)$

And we can use the complement rule like this

$P(X(2) >6) = 1-P(X(2)\leq 6)= 1-[P(X(2) =0)+P(X(2) =1)+P(X(2) =2)+...+P(X(2) =6)]$

And we can solve this like this using the masss function:

$P(X(2) >6) = 1- [e^{-4}+ \frac{e^{-4}4^1}{1!}+ \frac{e^{-4}4^2}{2!} +\frac{e^{-4}4^3}{3!} +\frac{e^{-4}4^4}{4!}+\frac{e^{-4}4^5}{5!}+\frac{e^{-4}4^6}{6!}]$

And we got: $P(X(2) >6) =1-0.889=0.111$

Part c

For this case we know that the arrival time follows an exponential distribution and let T the random variable:

$T \sim Exp(\lambda=2)$

The probability of no arrival during a period of duration t is given by:

$f(T) = e^{-\lambda t}$

And we want to find a value of t who satisfy this:

$e^{-2t} \leq 2$

We can apply natural log in both sides and we got:

$-2t \leq ln(0.2)$

If we multiply by -1 both sides of the inequality we have:

$2t \geq -ln(0.2)$

And if we divide both sides by 2 we got:

$t \geq \frac{-ln(0.2)}{2}$

$t \geq 0.8047$

And then we can conclude that the time period with any load would be 0.8047 years.

3 0

4 years ago

HELP 25pts and will mark brainliest. I NEED THIS NOW!! Which expression shows the result of applying the distributive property t

Arte-miy333 [17]

The last bottom problem

6 0

4 years ago

What is happening to the graph when the x-values are between 4 and 8

VashaNatasha [74]

The numbers are going down

5 0

3 years ago

Read 2 more answers

I want to know algebra

GarryVolchara [31]

Answer:

ssdfhhhgfhhfhbcfhvzsfb dgg

5 0

3 years ago