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mojhsa [17]
3 years ago
7

A 16 oz package of brown rice costs 79 cents and 32 oz package of white rice costs 3.49 which package is a better deal and why

Mathematics
1 answer:
Nataly [62]3 years ago
3 0
The better deal is the 16 ounces for 79 cents because it is 5 cents an ounce
while the 32 ounce bag is 11 cents an ounce
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Subtract the 2x and divide by -4. y=1/2x-2 or y=x/2-2
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Solve the equation cos (x/2) = cos x + 1. what are the solutions on the interval 0° ≤ x < 360°?
Sedbober [7]

Answer:

Step-by-step explanation:

cos (x/2)=cos x+1

cos (x/2)=2cos ²(x/2)

2 cos²(x/2)-cos (x/2)=0

cos (x/2)[2 cos (x/2)-1]=0

cos (x/2)=0=cos π/2,cos (3π/2)=cos (2nπ+π/2),cos(2nπ+3π/2)

x/2=2nπ+π/2,2nπ+3π/2

x=4nπ+π,4nπ+3π

n=0,1,2,...

x=π,3π

or x=180°,540°,...

180°∈[0,360]

so x=180°

or

2cos(x/2)-1=0

cos (x/2)=1/2=cos60,cos (360-60)=cos 60,cos 300=cos (360n+60),cos (360n+300)

x/2=360n+60,360n+300

x=720n+120,720n+300

n=0,1,2,...

x=120,300,840,1020,...

only 120° and 300° ∈[0,360°]

Hence x=120°,180°,300°

7 0
2 years ago
The answer and how to do it.
Kruka [31]

let x = orginal price of the shorts

$21 = x(100%-20%) * 1.05

$21 = x(80%) * 1.05

$21 = 0.8x * 1.05

Subtract 1.05 from both sides

$19.95 = 0.8x

Divide 0.8 from both sides

$24.9375 = x

So the orginal price of the shorts are about $24.94

4 0
3 years ago
Someone please help :(!!!!!!
dezoksy [38]

Answer:

GE

Step-by-step explanation:

6 0
3 years ago
Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????
LenaWriter [7]

Answer:

E[X^2]= \frac{2!}{2^1 1!}= 1

Var(X^2)= 3-(1)^2 =2

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

\phi(t) = E[e^{tX}]

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx

And we have that the moment generating function can be write like this:

\phi(t) = e^{\frac{t^2}{2}

And we can write this as an infinite series like this:

\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]

E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}

And then we have this:

E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...

And then we can find the E[X^2]

E[X^2]= \frac{2!}{2^1 1!}= 1

And we can find the variance like this :

Var(X^2) = E[X^4]-[E(X^2)]^2

And first we find:

E[X^4]= \frac{4!}{2^2 2!}= 3

And then the variance is given by:

Var(X^2)= 3-(1)^2 =2

7 0
3 years ago
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