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8_murik_8 [283]
3 years ago
8

Which of the following points is on a circle if its center is (-13,-12) and a point on the circumference is (-17, -12)?

Mathematics
1 answer:
tia_tia [17]3 years ago
4 0

Answer:

D

Step-by-step explanation:

Obtain the equation of the circle in standard form

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

here (h, k) = (- 13, - 12), thus

(x + 13)² + (y + 12)² = r²

The radius is the distance from the centre (- 13, - 12) to the point on the circumference (- 17, - 12)

Use the distance formula to calculate r

r = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 17, - 12) and (x₂, y₂ ) = (- 13, -12)

r = \sqrt{(-13+17)^2+(-12+12)^2} = \sqrt{16} = 4

Hence

(x + 13)² + (y + 12)² = 16 ← in standard form

Substitute the coordinates of each point into the left side of the equation and check

A (- 17, - 13) : (- 4)² + (- 1)² = 16 + 1 = 17 ≠ 16

B (- 9, - 17) : 4² + (- 5)² = 16 + 25 = 41 ≠ 16

C (- 12, 13) : 1² + 25² ≠ 16

D (- 9, - 12) : 4² + 0² = 16

Since (- 9, - 12) satisfies the equation, it is on the circle → D

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A sample of 200 observations from the first population indicated that x1 is 170. A sample of 150 observations from the second po
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Answer:

a) For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b) Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c)z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d) Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

Step-by-step explanation:

Data given and notation    

X_{1}=170 represent the number of people with the characteristic 1

X_{2}=110 represent the number of people with the characteristic 2  

n_{1}=200 sample 1 selected  

n_{2}=150 sample 2 selected  

p_{1}=\frac{170}{200}=0.85 represent the proportion estimated for the sample 1  

p_{2}=\frac{110}{150}=0.733 represent the proportion estimated for the sample 2  

\hat p represent the pooled estimate of p

z would represent the statistic (variable of interest)    

p_v represent the value for the test (variable of interest)  

\alpha=0.05 significance level given  

Concepts and formulas to use    

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:    

Null hypothesis:p_{1} = p_{2}    

Alternative hypothesis:p_{1} \neq p_{2}    

We need to apply a z test to compare proportions, and the statistic is given by:    

z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}   (1)  

a.State the decision rule.

For this case the value of the significanceis \alpha=0.05 and \alpha/2 =0.025, we need a value on the normal standard distribution thataccumulates 0.025 of the area on each tail and we got:

z_{\alpha/2} =1.96

If the calculated statistic |z_{calc}| >1.96 we can reject the null hypothesis at 5% of significance

b. Compute the pooled proportion.

Where \hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{170+110}{200+150}=0.8  

c. Compute the value of the test statistic.                                                                                              

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.    

Replacing in formula (1) the values obtained we got this:    

z=\frac{0.85-0.733}{\sqrt{0.8(1-0.8)(\frac{1}{200}+\frac{1}{150})}}=2.708    

d. What is your decision regarding the null hypothesis?

Since the calculated value satisfy this condition 2.708>1.96 we have enough evidence at 5% of significance that we have a significant difference between the two proportions analyzed.

5 0
3 years ago
Three listening stations located at (3300, 0), (3300, 1100), and (-3300, 0) monitor an explosion. The last two stations detect t
erastovalidia [21]

Answer:

The coordinates of the explosion is (3300, -2750)

Step-by-step explanation:

I have plot the points and attached to thus answer for easy understanding.

Now, from the question, since station A is the first to hear the explosion, we'll make it the foci of the parabola in the graph I attached and it will be horizontal since the distance between station C and A is much more than that between station B and A. Thus, the reason why station C will have to be the other foci with the hyperbola centred at the origin.

Now, sound travels at a speed of 1100 ft/s and station B is located 1100 ft from station A. Thus, the explosion would likely have occurred at a point on the line x = 3300ft . Since station A is 3300ft from centre C = 3300,hence C² = 3300² = 10,890,000. Since it takes 4 seconds longer for the sound to reach station C than A, the sound has traveled 4(1100)= 4400 ft.

Thus, 4400 = d1 = d2 = 2a

So,2a = 4400 and so, a =2200

a² = 2200² = 4,840,000 where d1 is the distance from station C to the explosion and d2 is the distance from station A to the explosion. To find b², let's use the equation ;

c² = a² + b² and so; b² = c² - a² = 10,890,000 - 4,840,000 = 6,050,000

Equation of hyperbola is given as;

(x²/a²) - (y²/b²) = 1

Plugging in the values of a² and b², we obtain ;

(x²/4,840,000) - (y²/6,050,000) = 1

Since we have deduced that the explosion must occur on the line x= 3300, we'll put in 3300 for x to obtain ;

(3300²/4,840,000) - (y²/6,050,000) = 1

2.25 - 1 = (y²/6,050,000)

y² = (1.25 x 6,050,000)

y² = 7562500

y = √7562500

y = ± 2750

Due to the fact that the explosion will occur at a point further from station B than from station A, the explosion will take place in quadrant 4. Thus, we will take the negative value of y which is - 2750.

So explosion will occur at the coordinate (3300, -2750)

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Answer:

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