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3 years ago

10

help please! Darren is finding the equation in the form y = m x + b for a trend line that passes through the points (2, 18) and

(–3, 8). Which value should he use as b in his equation? a) –34 b) –19 c) 2 d) 14

2 answers:

andrezito [222]3 years ago

6 0

Answer: d) 14

Step-by-step explanation:

Equation of a line passing through (a,b) and (c,d):

$(y-b)=\dfrac{d-b}{c-a}(x-a)$

Equation of a line passing through (2, 18) and (–3, 8):

$(y-18)=\dfrac{8-18}{-3-2}(x-2)\\\\\Rightarrow\ (y-18)=\dfrac{-10}{-5}(x-2)\\\\\Rightarrow\ (y-18)=2(x-2)\\\\\Rightarrow\ y-18=2x-4\\\\\Rightarrow\ y=2x-4+18\\\\\Rightarrow\ y=2x+14$

Comparing resulting equation $y=2x+14$ to $y = m x + b$ , we get value of b= 14.

Hence, correct option is d) 14

Rudiy273 years ago

6 0

Answer:

D

Step-by-step explanation:

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The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f

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Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

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n₂= 1000

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The formula for the interval is:

p'₁± $Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }$

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0.37±1.648* $\sqrt{\frac{0.37*0.63}{1000} }$

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With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂± $Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }$

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0.48±1.648*0.016

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With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

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Seniors: $\sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016$

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

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bezimeni [28]

Answer:

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Step-by-step explanation:

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