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solmaris [256]
3 years ago
11

The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 oun

ces and a standard deviation of 0.30 ounce. The company receives complaints from consumers who actually measure the amount of soda in the cans and claim that the volume is less than the advertised 12 ounces. What proportion of the soda cans contain less than the advertised 12 ounces of soda?
Mathematics
1 answer:
lorasvet [3.4K]3 years ago
6 0

Answer: 0.0668

Step-by-step explanation:

Given : The volume of soda a dispensing machine pours into a 12-ounce can of soda follows a normal distribution with a mean of 12.45 ounces and a standard deviation of 0.30 ounce.

i.e. \mu=12.45 and \sigma=0.30

Let x denotes the volume of soda a dispensing machine pours into a 12-ounce can.

Then, the proportion of the soda cans contain less than the advertised 12 ounces of soda will be :-

P(x

Hence, the proportion of the soda cans contain less than the advertised 12 ounces of soda = 0.0668

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A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the other three sides
givi [52]

Answer:

x  =  28 m

y  =  14  m

A(max)  =  392 m²

Step-by-step explanation:

Rectangular garden    A (r ) =  x * y

Let´s call x the side of the rectangle to be constructed with a rock wall, then only one x side of the rectangle will be fencing with wire.

the perimeter of the rectangle is  p  =  2*x  +  2*y    ( but in this particular case only one side x will be fencing with wire

56   =   x    +  2*y      56   -  2*y  =  x

A(r)   =  ( 56  -  2*y ) * y

A(y ) =  56*y  -  2*y²

Tacking derivatives on both sides of the equation we get:

A´(y )  =  56  - 4 * y        A´(y) = 0     56  -  4*y  =  0    4*y  =  56

y =  14 m

and x  =  56  - 2*y    =  56 - 28  = 28 m

Then dimensions of the garden:

x  =  28 m

y  =  14  m

A(max)  =  392 m²

How do we know that the area we found is a local maximum??

We find the second derivative

A´´(y)  = - 4     A´´(y)  <  0   then the function A(y) has a local maximum at y = 14 m

4 0
3 years ago
Find the inverse of the function.<br> y= √x - 4
EastWind [94]

Answer:

The inverse is f^{-1} (x) = (x+4)^{2}

Step-by-step explanation:

(By the way, f^{-1}(x) is f negative one of x, like f(x), but inverse)

f(x) = \sqrt{x} - 4 is the same as y = \sqrt{x} - 4

y = \sqrt{x} - 4               start with the equation.

x = \sqrt{y} - 4               switch x and y.

x + 4 = \sqrt{y}              add 4 to both sides.

(x+4)^{2} = y             square both sides.

f^{-1} (x) = (x+4)^{2}    change y to the inverse f(x) format.

I hope this helps! :)

4 0
3 years ago
Does anyone know the answers to this
sveticcg [70]

Part A:

10p + 12v > 3200

p > 12000

v < 15000

(imagine they have the line underneath the symbols)

Part B:

at least 14,000 cups of plain yogurt

(make sure to convert the dollars to cents so that it is accurate)

math:

10p + 12(15000) > 320000

10p > 140000

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5 0
3 years ago
Given that PS is the perpendicular bisector of QR, PQ=3.5m+18, and PR=6m+3, identify PQ.
kolbaska11 [484]

Answer:

PQ=39\ units

Step-by-step explanation:

we know that

QS=RS ----> equation A

Applying Pythagoras Theorem

PQ^2=SP^2+QS^2 -----> equation B

PR^2=SP^2+RS^2-----> equation C

substitute equation A in equation C

PR^2=SP^2+QS^2

we have that

PR^2=PQ^2

so

PR=PQ

substitute the given values

6m+3=3.5m+18

Solve for m

6m-3.5m=18-3

2.5m=15

m=6

Find the value of PQ

PQ=3.5m+18

substitute the value of m

PQ=3.5(6)+18

PQ=39\ units

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The answer should be 3 :)

8 0
3 years ago
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