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Lady bird [3.3K]
3 years ago
11

a country club offers membership to it's customers. there is a one-time application fee of $75 and the dues are $25 per month. w

rite an equation to represent the total cost of membership at this country club​
Mathematics
1 answer:
Lilit [14]3 years ago
3 0

Answer: 25x + 75

Step-by-step explanation:

The 25 is the monthly fee, x stands for the number of months and the 75 is the set fee

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Solve the equation for x. 1 6 (2x − 12x) = 30 A) −12 B) −15 C) −18 D) −24
lubasha [3.4K]

\frac{1}{6}*(2x-12x) = 30

\frac{1}{6}*(-10x) = 30

\frac{-10}{6}x = 30

\frac{-5}{3}x = 30

\frac{5}{3}x = -30

x = -30/\frac{5}{3} = -30 * \frac{3}{5}

x = -18

the correct answer is C

8 0
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An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side. 
The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day. 
7 0
3 years ago
W(t) = 3t – 1; t = 5
son4ous [18]
I’m not sure if your asking for the solution, but Hope this helps!

8 0
3 years ago
A poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ag
Katyanochek1 [597]

Answer:

At 5% significance level, it is statistically evident that    there is nodifference in the proportion of college students who consider themselves overweight between the two poll                                  

Step-by-step explanation:

Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.

Let five years ago be group I X and as of now be group II Y

H_0: p_x =p_y\\H_a: p_x \neq p_y

(Two tailed test at 5% level of significance)

                Group I            Group II              combined p

n                  270                 300                         570

favor            120                  140                          260

p                   0.4444            0.4667                   0.4561

Std error   for differene = \sqrt{\frac{0.4561(1-0.4561)}{570} } \\=0.0209

p difference = -0.0223    

Z statistic = p diff/std error =       -1.066

p value =0.2864

Since p value >0.05, we accept null hypothesis.

        At 5% significance level, it is statistically evident that    there is nodifference in the proportion of college students who consider themselves overweight between the two poll                                  

7 0
3 years ago
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