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alexandr402 [8]
2 years ago
7

The amount of water dispensed by a water dispenser is normally distributed, with a mean of 12.10 ounces and a standard deviation

of 0.25 ounces. In which range will the amount of water dispensed be found 99.7% of the time?
Mathematics
1 answer:
rjkz [21]2 years ago
8 0

Answer:

11.35 ounce and 12.85 ounce

Step-by-step explanation:

According to the Empirial rule of the distribution under the normal curve falls within 3 standard deviations of the mean.

That is:

\mu \pm3 \sigma

From the given information, the mean is

\mu = 12.10

and the standard deviation is

\sigma = 0.25

We substitute the given parameters to obtain;

12.10 \pm3(0.25)

12.10 \pm0.75

This means the lower limit is

12.10 -0 .75 = 11.35

and the upper limit is

12.10 + 0.75 = 12.85

Therefore 99.7% of the amount of water dispensed is between 11.35 ounce and 12.85 ounce

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A waste management company is designing a rectangular construction dumpster that will be twice as long as it is wide and must ho
photoshop1234 [79]

Answer:

The dimensions that minimize the surface are:

Wide: 1.65 yd

Long: 3.30 yd

Height: 2.20 yd

Step-by-step explanation:

We have a rectangular base, that its twice as long as it is wide.

It must hold 12 yd^3 of debris.

We have to minimize the surface area, subjet to the restriction of volume (12 yd^3).

The surface is equal to:

S=2(w*h+w*2w+2wh)=2(3wh+2w^2)

The volume restriction is:

V=w*2w*h=2w^2h=12\\\\h=\frac{6}{w^2}

If we replace h in the surface equation, we have:

S=2(3wh+2w^2)=6w(\frac{6}{w^2})+4w^2=36w^{-1}+4w^2

To optimize, we derive and equal to zero:

dS/dw=36(-1)w^{-2} + 8w=0\\\\36w^{-2}=8w\\\\w^3=36/8=4.5\\\\w=\sqrt[3]{4.5} =1.65

Then, the height h is:

h=6/w^2=6/(1.65^2)=6/2.7225=2.2

The dimensions that minimize the surface are:

Wide: 1.65 yd

Long: 3.30 yd

Height: 2.20 yd

8 0
2 years ago
Charlotte’s weekly paycheck is based on the number of hours worked during the week and on the weekend. If she works 13 hours dur
Gelneren [198K]
I am setting the week hourly rate to x, and the weekend to y.  Here is how the equation is set up:

13x + 14y = $250.90
15x + 8y = $204.70

This is a system of equations, and we can solve it by multiplying the top equation by 4, and the bottom equation by -7.  Now it equals:

52x + 56y = $1003.60
-105x - 56y = -$1432.90

Now we add these two equations together to get:

-53x = -$429.30 --> 53x = $429.30 --> (divide both sides by 53) x = 8.10.  This is how much she makes per hour on a week day.

Now we can plug in our answer for x to find y.  I am going to use the first equation, but you could use either.

$105.30 + 14y = $250.90.  Subtract $105.30 from both sides --> 14y = $145.60 divide by 14 --> y = $10.40

Now we know that she makes $8.10 per hour on the week days, and $10.40 per hour on the weekends.  Subtracting 8.1 from 10.4, we figure out that she makes $2.30 more per hour on the weekends than week days.
7 0
2 years ago
Read 2 more answers
Can I get some help plz​
Mila [183]

Step-by-step explanation:

In ️ BEA and ️ CED

EB=EC ----given

angle ABE= angle DCE --------- given

angle DEB= angle DCE -------vertically opposite angles are equal

Hence, ️ BEA is congruent to ️ CED

3 0
2 years ago
Equivalent expression for 12+18n+7-14n
elena55 [62]
19+4n is the answer and i m sure
3 0
2 years ago
Read 2 more answers
The 3rd degree Taylor polynomial for cos(x) centered at a = π 2 is given by, cos(x) = − (x − π/2) + 1/6 (x − π/2)3 + R3(x). Usin
Otrada [13]

Answer:

The cosine of 86º is approximately 0.06976.

Step-by-step explanation:

The third degree Taylor polynomial for the cosine function centered at a = \frac{\pi}{2} is:

\cos x \approx -\left(x-\frac{\pi}{2} \right)+\frac{1}{6}\cdot \left(x-\frac{\pi}{2} \right)^{3}

The value of 86º in radians is:

86^{\circ} = \frac{86^{\circ}}{180^{\circ}}\times \pi

86^{\circ} = \frac{43}{90}\pi\,rad

Then, the cosine of 86º is:

\cos 86^{\circ} \approx -\left(\frac{43}{90}\pi-\frac{\pi}{2}\right)+\frac{1}{6}\cdot \left(\frac{43}{90}\pi-\frac{\pi}{2}\right)^{3}

\cos 86^{\circ} \approx 0.06976

The cosine of 86º is approximately 0.06976.

8 0
2 years ago
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