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GarryVolchara [31]

1 year ago

13

Binks needs to raise at least $3600. He was given $200 and plans to save $50 a week. Select all of the inequalities that represe

nt this situation where, n, is the number of weeks.
50n≤3400

50n−200≤3600

50n+200≤3600

50n≥3400

50n−200≥3600

50n+200≥3600

1 answer:

kherson [118]1 year ago

5 0

Answer:

I'm pretty sure it is 50n+200 <u><</u> 3600

Step-by-step explanation:

Im to fully sure tho

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Let e1= 1 0 and e2= 0 1 , y1= 4 5 , and y2= −2 7 , and let T: ℝ2→ℝ2 be a linear transformation that maps e1 into y1 and maps

Furkat [3]

Answer:

The image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is $\left[\begin{array}{c}24&-8\end{array}\right]$

Step-by-step explanation:

We know that $T:$ $IR^{2}$ → $IR^{2}$ is a linear transformation that maps $e_{1}$ into $y_{1}$ ⇒

$T(e_{1})=y_{1}$

And also maps $e_{2}$ into $y_{2}$ ⇒

$T(e_{2})=y_{2}$

We need to find the image of the vector $\left[\begin{array}{c}4&-4\end{array}\right]$

We know that exists a matrix A from $IR^{2x2}$ (because of how T was defined) such that :

$T(x)=Ax$ for all x ∈ $IR^{2}$

We can find the matrix A by applying T to a base of the domain ( $IR^{2}$ ).

Notice that we have that data :

$B_{IR^{2}}=$ { $e_{1},e_{2}$ }

Being $B_{IR^{2}}$ the cannonic base of $IR^{2}$

The following step is to put the images from the vectors of the base into the columns of the new matrix A :

$T(\left[\begin{array}{c}1&0\end{array}\right])=\left[\begin{array}{c}4&5\end{array}\right]$ (Data of the problem)

$T(\left[\begin{array}{c}0&1\end{array}\right])=\left[\begin{array}{c}-2&7\end{array}\right]$ (Data of the problem)

Writing the matrix A :

$A=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]$

Now with the matrix A we can find the image of $\left[\begin{array}{c}4&-4\\\end{array}\right]$ such as :

$T(x)=Ax$ ⇒

$T(\left[\begin{array}{c}4&-4\end{array}\right])=\left[\begin{array}{cc}4&-2\\5&7\\\end{array}\right]\left[\begin{array}{c}4&-4\end{array}\right]=\left[\begin{array}{c}24&-8\end{array}\right]$

We found out that the image of $\left[\begin{array}{c}4&-4\end{array}\right]$ through T is the vector $\left[\begin{array}{c}24&-8\end{array}\right]$

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1 year ago

A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

<u>Rate Out</u>

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

$20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000$

Therefore, the amount of salt in the tank at any time t is:

$A(t)=(50+t)-75000(50+t)^{-2}$

After 15 minutes, the amount of salt in the tank is:

$A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds$

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1 year ago

A quality control officer randomly checking weights of pumpkin seed bags being filled by an automatic filling machine. Each bag

GarryVolchara [31]

From 0g to 494.7g are rejected as well as those heavier than 505.3grams.

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7 0

2 years ago

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saul85 [17]

Answer:

0.99865

Step-by-step explanation:

The question above is modelled by gaussian distribution. Gaussian distribution is also known as Normal distribution.

To solve the above question, we would be using the z score formula

The formula for calculating a z-score

z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

In the above question,

x is 115 liters

μ is 100

σ is the population standard deviation is unknown. But we were given variance in the question.

Standard deviation = √Variance

Variance = 9

Hence, Standard deviation = √9 = 3

We go ahead to calculate our z score

z = (x-μ)/σ

z = (115 - 100) / 3

z = 15/ 3

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Using the z score table of normal distribution to find the Probability of having a z score of 5

P(x = 115) = P(z = 5) =

0.99865

Therefore the probability that this year it will produce 115 liters of wine = 0.99865

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vampirchik [111]

Answer:

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Step-by-step explanation:

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