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3 years ago

NEED THE ANSWER ASAP PLEASE

Mathematics

1 answer:

Irina-Kira [14]3 years ago

7 0

It's a linear program; the extrema will be at the corners.

Let's enumerate them. 4 choose 2 gives 6 possible meets of 4 lines. One pair are parallel, down to five to try.

4x-y=1 intersects x=0 at y=-1, outside the domain y≥0.

4x-y=1 intersects y=0 at x=1/4, (1/4, 0)

4x-y=1 intersects x=5 at y=19, (5,19)

(0,0) is outside the domain 4(0)-0=0 which isn't ≥1.

(5,0) is a valid corner.

It's a triangular domain. Three points to try,

C(1/4,0) = 6(1/4) + 2(0) = 3/2

C(5,19) = 6(5) + 2(19) = 68

C(5,0) = 6(5) + 2(0) = 30

Answer: Maximum C=68 at (x,y)=(5,19)

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Step-by-step explanation:

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3 years ago

Folds - Circle Cliffs, UT. Both monoclines and anticlines uplift and deform rock layers such that when eroded, certain ages of r

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Answer:the relative ages of the rocks exposed in the circle cliff area are given below.

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2. younger rocks are exposed in the center and older rocks in the flanking flatirons.

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2. But when younger rocks are exposed in contrast to the older rocks, these younger rocks are exposed at the center while the older ones receive exposure at the flanking flatirons.

Note that both cases interchange, exposure of a particular rock occur at the center and the next category of rock receive theirs at flanking flatirons.

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3 years ago

A whale travel a distance of about 5000 miles each way at an average rate of 91 miles per day. About how many days does it take

Sphinxa [80]

It takes about 55 days.

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3 years ago

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3 years ago

Read 2 more answers

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 55 o

Usimov [2.4K]

Answer:

a) For this case and using the empirical rule we can find the limits in order to have 9% of the values:

$\mu -2\sigma = 55 -2*6 =43$

$\mu +2\sigma = 55 +2*6 =67$

95% of the widget weights lie between 43 and 67

b) For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

$100 -0.15-2.5 = 97.85$

c) We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

Step-by-step explanation:

For this case our random variable of interest for the weights is bell shaped and we know the following parameters.

$\mu = 55, \sigma =6$

We can see the illustration of the curve in the figure attached. We need to remember that from the empirical rule we have 68% of the values within one deviation from the mean, 95% of the data within 2 deviations and 99.7% of the values within 3 deviations from the mean.

Part a

For this case and using the empirical rule we can find the limits in order to have 9% of the values:

$\mu -2\sigma = 55 -2*6 =43$

$\mu +2\sigma = 55 +2*6 =67$

95% of the widget weights lie between 43 and 67

Part b

For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

$100 -0.15-2.5 = 97.85$

Part c

We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

4 0

4 years ago