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goblinko [34]
3 years ago
5

How many sports events would a student have to attend to make the pasa a better deal

Mathematics
1 answer:
Alexxandr [17]3 years ago
4 0
You would have to attend 8 events
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:) PLEASE!!!! HELP! some food while you solve <3
Dominik [7]

Answer:

1/3

Step-by-step explanation:

rise over run. Find two places that land on a line I picked (0,0) and (3,1) then count up one from the first point and over 3 for the second.

6 0
2 years ago
Read 2 more answers
HELP ASSSAP WITH THIS QUESTION
bulgar [2K]
The answer is linear and increasing. look at it.... it goes up. 

8 0
3 years ago
Please explain how to do this, I'm using all my points<br><br>A. 40 L<br>B. 80 L<br>C. 4 L <br>D. 8L
exis [7]

Answer: choice A. 40 liters

====================================

Explanation:

x = number of liters of the 50% alcohol solution

If we have x liters of 50% alcohol, then we have 0.50*x liters of pure alcohol. This is added to 0.90*40 = 36 liters of pure alcohol (from the 90% solution).

So far we have 0.50*x + 36. This expression represents the total amount of pure alcohol. We want a 70% solution, so we want 70% of the total 40+x meaning 0.50*x + 36 is to be set equal to 0.70*(40+x) and we solve for x as shown below

0.50*x + 36 = 0.70*(40+x)

0.50*x + 36 = 0.70*(40)+0.70*(x)

0.50*x + 36 = 28+0.70*x

36 - 28 = 0.70*x - 0.50x

8 = 0.20x

0.20x = 8

x = 8/0.20

x = 40

So that is why the answer is choice A. 40 liters

6 0
2 years ago
Write an equation, in slope-intercept form, to model each situation. You rent a bicycle for $20 plus $2 per hour.​
Leona [35]
Y=2x+20 x represents hours
3 0
3 years ago
The average monthly cell phone bill was reported to be $50.07 by the U.S. Wireless Industry. Random Sampling of a large cell pho
zalisa [80]

Answer:

The null and alternative hypotheses are:

H_{0}:\mu= 50.07

H_{a}:\mu>50.07

Under the null hypothesis, the test statistic is:

t=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}} }

Where:

\bar{x} = 52.86 is the sample mean

s=7.0132 is the sample standard deviation

n=10 is the sample size

\therefore t=\frac{52.86-50.07}{\frac{7.0132}{\sqrt{10}} }    

          =1.26

Now, the right tailed t critical value at 0.05 significance level for df = n-1 = 10-1 = 9 is:

t_{critical}=1.833

Since the t statistic is less than the t critical value at 0.05 significance level, therefore,we fail to reject the null hypothesis and conclude that there is not sufficient evidence to support the claim that the average phone bill has increased.


4 0
3 years ago
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