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Inga [223]
3 years ago
11

What is the equation of the line that passes through the point (4,-7) and has a

Mathematics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

y = -\frac{1}{2} x -5

point slope form....

y - (-7) = -1/2(x-4)

y + 7 = -x/2 + 2

2y + 14 = -x + 4

2y = -x -10

y = -1/2 x - 5

Step-by-step explanation:

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PLEASE HElp asap <br> Solve for x.<br> x = [?]<br> X + 19<br> 7x + 9
Bess [88]
X + 19 + 7x + 9 = 180
8x + 28 = 180
8x = 180 - 28
8x = 152
X = 152 : 8
X = 19
3 0
3 years ago
Which of the following best describes the slope of the line below?
strojnjashka [21]

Answer:

D. Undefined

Step-by-step explanation:

It is a vertical line so it has no slope.

5 0
3 years ago
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Solve for x.<br> JKL<br> 2x-4<br> 8<br> 16<br> 7<br> 56
mariarad [96]

you have to use intercept theorem, also known as Thales's theorem to slove this problem

7 0
1 year ago
Enter the coordinates of the point on the unit circle at the given angle. 150 degrees. please help!
Katarina [22]

Answer:

\boxed{(-\frac{\sqrt{3}}{2}, \frac{1}{2})}

Step-by-step explanation:

Method 1: Using a calculator <em>instead</em> of the unit circle

The unit circle gives coordinates pairs for the <em>cos</em> and <em>sin</em> values at a certain angle. Therefore, if an angle is given, use a calculator to evaluate the functions at cos(angle) and sin(angle).

Method 2: Using the unit circle

Use the unit circle to locate the angle measure of 150° (or 5π/6 radians) and use the coordinate pair listed by the value (see attachment).

This coordinate pair is (-√3/2, 1/2).

Download pdf
8 0
3 years ago
Read 2 more answers
The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f
Sloan [31]

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} }

0.37±1.648*0.015

[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
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