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Katen [24]
2 years ago
9

Tara is graphing the equation 4x + 2y = 10. Which of these shows the correct equation in slope-intercept form, slope, and y-inte

rcept? Question 2 options: y = –2x + 5 slope = –2 y-intercept = 5 y = 4x + 5 slope = 4 y-intercept = 5 y = 2x + 4 slope = 2 y-intercept = 4 y = –4x + 5 slope = –4 y-intercept = 5
Mathematics
1 answer:
blagie [28]2 years ago
5 0

the answer should be : y=-2x+5

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Let C represent the cost in dollars and R represent the revenue in dollars. What is the break-even point? Use a table to help if
BigorU [14]

Answer:

x = 5

R = C = 225

Step-by-step explanation:

Given

C = 15x + 150

R = 45x

Required

The break even point

To do this, we make use of:

R =C

45x=15x + 150

Collect Like Terms

45x-15x = 150

30x = 150

x = 5

Substitute 5 for x in R = 45x

R = 45 * 5

R = 225

Hence, the break even point is:

x = 5

R = C = 225

5 0
2 years ago
The two squares x and y are mathematically similar. The areas of x and y are 17cm squared and 272cm squared, respectively. The l
Rasek [7]
I would set this up as a a proportion:
\frac{17cm}{5cm}  · \frac{272cm}{x} 
17x = 1360
x = 20 cm
6 0
3 years ago
Answer the following.
nexus9112 [7]

Step-by-step explanation:

1075 students were asked how close they live to the college. The table shows the results below with some values missing. Enter the correct values in the boxes remembering to use the % symbol where necessary.

<u>20 to 25: </u>given 20%

20% of 1,075 = 215

<u>15 to 19: </u>given 12%

12% of 1,075 = 129

<u>10 to 14: </u>given 516 students

(516/1,075)*100 = 48%

<u>5 to 9: </u>given 162 students

(162/1,075)*100 = 15.07%

<u>< 5: </u>given 53 students

(53/1,075)*100 = 4.93%

3 0
1 year ago
4
Gala2k [10]

Answer:

yhjhgnxfvhjmnfdtjnnbnjyu

4 0
3 years ago
The weight of baby goats is believed to be Normally distributed, with a mean of 5.75 pounds. The average weight of a random samp
Julli [10]

Answer:

The standard error of the mean is 0.0783.

Step-by-step explanation:

The Central Limit Theorem helps us find the standard error of the mean:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

The standard deviation of the sample is the same as the standard error of the mean. So

SE_{M} = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\sigma = 0.35, n = 20

So

SE_{M} = \frac{\sigma}{\sqrt{n}}

SE_{M} = \frac{0.35}{\sqrt{20}}

SE_{M} = 0.0783

The standard error of the mean is 0.0783.

7 0
3 years ago
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