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yuradex [85]
3 years ago
8

The diameter of a circle has endpoints P(-12, -4) and Q(6, 12).

Mathematics
1 answer:
nexus9112 [7]3 years ago
3 0

9514 1404 393

Answer:

  (x +3)² +(y -4)² = 145

Step-by-step explanation:

The center of the circle is the midpoint of the given segment PQ. If we call that point A, then ...

  A = (P +Q)/2

  A = ((-12, -4) +(6, 12))/2 = (-12+6, -4+12)/2 = (-6, 8)/2

  A = (-3, 4)

The equation of the circle for some radius r is ...

  (x -(-3))² +(y -4)² = r² . . . . . . where (-3, 4) is the center of the circle

The value of r² can be found by substituting either of the points on the circle. If we use Q, then we have ...

  (6 +3)² +(12 -4)² = r² = 9² +8²

  r² = 81 +64 = 145

Then the equation of the circle is ...

  (x +3)² +(y -4)² = 145

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Answer:

D. y = -1/4x + 3

Step-by-step explanation:

If the line is perpendicular, it will have a opposite reciprocal slope.

So, the slope will be -1/4.

Plug in this slope and the given point into y = mx + b, and solve for b:

y = mx + b

-6 = -1/4(12) + b

-6 = -3 + b

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Plug in the slope and y intercept into y = mx + b

y = -1/4x - 3

So, the correct answer is D. y = -1/4x + 3

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3 years ago
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MrRissso [65]
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3 years ago
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A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi
34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{X}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{X}=\sqrt{n}\sigma

It is provided that:

\mu=\$27\\\sigma=\$18\\n=310

As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

5 0
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You can round 976 to 1000 and 522 to 500. So it would be
1000 - 500 = 500. The answer you would get after subtracting 976-522 would be more than 400 if you round it.
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Sav [38]
404 is the answer. if you add them all together
8 0
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