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horrorfan [7]
3 years ago
5

Simultaneous Equation (x+5)(x+2)=65 , xy=24 Plssss help me ✌

Mathematics
1 answer:
Ne4ueva [31]3 years ago
7 0

Answer:

  • (-11.70, -2.05), (4.70, 5.10)

Step-by-step explanation:

<u>Given equations</u>

  • (x+5)(x+2)=65
  • xy=24

<u>Solving the first equation for x:</u>

  • (x+5)(x+2)=65
  • x^2 + 7x + 10 = 65
  • x^2 + 7x - 55 = 0
  • x = (-7 ± √(49+4*55))/2
  • x = -11.70, x = 4.70

<u>Then solving for y:</u>

  • y= 24/(-11.7) = -2.05
  • y = 24/4.7 = 5.10

<u>Solution set: </u>

  • (-11.70, -2.05), (4.70, 5.10)
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2 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
3 years ago
488, 460, 520, 544, 535<br> What is the range of the data?
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Answer:

84

Step-by-step explanation:

To find the range, find the difference between the largest value and the smallest value.

544 - 460 = 84

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