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Alinara [238K]
3 years ago
11

The midpoint of (2,7) and (6,3)

Mathematics
1 answer:
White raven [17]3 years ago
8 0

Answer:

3

Step-by-step explanation:

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What are the units? Tell me the units so I can solve this please.

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The sea ice area around the North Pole fluctuates between about 6 million square kilometers in September to 14 million square ki
Aleksandr [31]

Answer:

there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

Step-by-step explanation:

Given the data in the question;

Let S(t) represent the amount sea ice around the North Pole in millions of square meters at a given time t,

t is the number of months since January.

Now, we use a cosine curve to model this scenario

Vertical shift will be;

D = ( 6 + 14 ) / 2 = 20 / 2

D = 10

Next is the Amplitude;

|A| = ( 6 - 14 ) / 2

|A| = 4

Now, the horizontal stretch factor will be;

B = 2π / 12

B = π/6

Hence;

S(t) = 4cos( π/6 × t ) + 10 ----------- let this be equation 1

Now we find when there will be less than 9 million square meters of sea ice;

S(t) = 9

so we have

9 = 4cos( π/6 × (t-2) ) + 10

9 - 10 = 4cos( π/6 × (t-2) )

-1 = 4cos( π/6 × (t-2) )

-1/4 = cos( π/6 × (t-2) )

so we have;

cos⁻¹( -1/4 ) = π/6 × (t₁-2)  -------- let this be equation 2

2π - cos⁻¹( -1/4 ) = π/6 × (t₂-2)  -------- let this be equation 3

so we solve equation 2 and 3

we have'

t₁ - t₂ = 6/π × ( 2π - cos⁻¹( -1/4 ) - cos⁻¹( -1/4 ) )

t₁ - t₂ = 6/π × ( 2π - 2cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - cos⁻¹( -1/4 )  

t₁ - t₂ = 6/π × ( π - 104.4775 )

t₁ - t₂ = 6/π × ( π - 104.4775 )  

t₁ - t₂ = 5.035

therefore, there are approximately 5.035 months when there is less than 9 million square meters of sea ice around the North Pole in a year.

6 0
3 years ago
Strawberries are $3.99 per pound. If a customer bought $30 worth, how many pounds did he buy? $
lubasha [3.4K]

Answer:

7.5

Step-by-step explanation:

Divide 30 by 3.99

  • 30/3.99=7.5
3 0
3 years ago
It is generally believed that the mean value of the district's total SAT score distribution is equal to 1200 (the null hypothesi
g100num [7]

Answer:

The corresponding p-value, is p = 1

Step-by-step explanation:

The maximum score SAT score, n =  1,600

The mean of the district's total SAT score distribution = 1,200

The claim of one of the districts principal, is the that mean of the district's total SAT score distribution ≠ 1,200

Using proportions, we have;

p = 1,200/1,600 = 0.75

q = 1 - p = 0.25

The margin of error, E = Z√(p·q/n)

∴ E = 5% = Z×√((0.75 × 0.25)/1,600)

z = 0.05/(√((0.75 × 0.25)/1,600)) ≈ 4.61880

Therefore, the corresponding p-value, p = 1

7 0
3 years ago
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