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ElenaW [278]
3 years ago
6

His is the table of values for y = 3x - 1 Can someone help me complete it, please.

Mathematics
2 answers:
avanturin [10]3 years ago
7 0

Answer:

1st blank :

-4

star :

5

it's correct answers

motikmotik3 years ago
4 0

Answer:

For the first column, the y= -4

For the last column, y= 5

Step-by-step explanation:

For the first column explanation:

  • y=3x-1
  • (plug in values) ---> 3(-1)-1
  • =-3-1
  • =-4
  • Check!
  • -4=3(-1)-1
  • -4 = -4 ✅

For the last column explanation:

  • y=3x-1
  • (plug in values) ---> 3(2)-1
  • =6-1
  • =5
  • Check!
  • 5=3(2)-1
  • 5=5 ✅

I hope this helps!

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3 years ago
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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Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

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

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen  who carry a credit card balance.

The formula for the interval is:

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

Z_{1-\alpha /2}= Z_{0.95}= 1.648

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

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[0.35;0.39]

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} }

0.48±1.648* \sqrt{\frac{0.48*0.52}{1000} }

0.48±1.648*0.016

[0.45;0.51]

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.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: \sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015

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.

8 0
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Answer:

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