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Karolina [17]
3 years ago
5

0.2x + 0.3y = 0.5 need help plzz​

Mathematics
1 answer:
krok68 [10]3 years ago
6 0

Answer:

y = − 0.  6 x + 1.6

Step-by-step explanation:

Write in slope-intercept form,  

y = m x + b

Note:

  • The decimals (0.6) and (1.6) ARE repeating...

If you need any other form of working it out let me know!

Glad I could help!!

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Sally goes out to lunch and has the option
castortr0y [4]

Hello,

well, that depend on what your asking. comment the answer choices so I can answer this question for you!

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3 0
3 years ago
Simplify the expression (4x − 3)(x + 5). 4x2 − 17x + 15 4x2 − 17x − 15 4x2 + 17x + 15 4x2 + 17x − 15
DanielleElmas [232]
There are mnemonic acronyms out there for reminding you how to do this. I think of it simply as using the distributive property.
  (4x-3)(x+5) = 4x(x+5) -3(x+5)
then again
  = (4x² +20x) + (-3x -15)
  = 4x² + (20-3)x -15 . . . . . . . combine like terms
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5 0
3 years ago
Read 2 more answers
An important factor in selling a residential property is the number of times real estate agents show a home. A sample of 23 home
yanalaym [24]

Using the t-distribution, it is found that:

a. The <u>margin of error</u> is of 4.7 homes.

b. The 98% confidence interval for the population mean is (19.3, 28.7).

The information given in the text is:

  • Sample mean of \overline{x} = 24.
  • Sample standard deviation of s = 9.
  • Sample size of n = 23.

We are given the <u>standard deviation for the sample</u>, which is why the t-distribution is used to solve this question.

The confidence interval is:

\overline{x} \pm M

The margin of error is:

M = t\frac{s}{\sqrt{n}}

Item a:

The critical value, using a t-distribution calculator, for a two-tailed <u>98% confidence interval</u>, with 23 - 1 = <u>22 df</u>, is t = 2.508.

Then, the <em>margin of error</em> is:

M = t\frac{s}{\sqrt{n}} = 2.508\frac{9}{\sqrt{23}} = 4.7

Item b:

The interval is:

\overline{x} - M = 24 - 4.7 = 19.3

\overline{x} + M = 24 + 4.7 = 28.7

The 98% confidence interval for the population mean is (19.3, 28.7).

A similar problem is given at brainly.com/question/15180581

3 0
2 years ago
Which statement is true about the purpose of trend lines in scatterplots? A trend line must pass through the largest data point.
Eddi Din [679]

Answer:

a trend line can be used to make predictions in real-world situations (the third option)

Explanation:

the first & second options are wrong because it is not a requirement for a trend line to pass through any given data point.

the fourth option is wrong because a trend line shows the general upward/downward trend of the data, and may not necessarily touch all of the data points.

the third option is correct because a trend line helps to roughly estimate the trend (hence the name) of the graph, which allows us to make predictions in real-world situations.

i hope this helps! :D

3 0
3 years ago
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Solve for t.
saw5 [17]

Answer:

1/3 is the answer I got I'm not sure if the top answer is a typo but that's the answer I got

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3 years ago
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