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Lisa [10]
3 years ago
5

What is the slope of a trend line that passes through the points (1, 3) and (10, 25)?​

Mathematics
1 answer:
Rama09 [41]3 years ago
4 0

Answer:

22 / 9

Step-by-step explanation:

25 - 3 = 22

10 - 1 = 9

22 / 9

You cannot simplify 22 / 9

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What is the connection between ratios, fractions, and percents?
devlian [24]
Fractions and percents are simply.different ways to express a ratio between two numbers.
3 0
3 years ago
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Solve the following...<br>6x - 5 = -9x -9​
sergiy2304 [10]

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:x = -\cfrac{4}{15}

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \: 6x - 5 =  - 9x - 9

[ add 9x and 5 on both sides ]

\qquad \tt \rightarrow \: 6x - 5 + 9x + 5 =  - 9x - 9 + 9x + 5

\qquad \tt \rightarrow \: 6x + 9x =  - 9 + 5

\qquad \tt \rightarrow \: 15x = -  4

\qquad \tt \rightarrow \: x =  -  \cfrac{4}{15}

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

7 0
2 years ago
Read 2 more answers
If B has the points, (-2,0) and (3,4), use the point slope formula to calculate the equation of the line.
pogonyaev
5/4 would be ur answer point slip formula is y^2 - y^1 over x^2 - x^1
5 0
3 years ago
A simple random sample of 90 is drawn from a normally distributed population, and the mean is found to be 138, with a standard d
bagirrra123 [75]

The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>

Suppose that we have:

  • Sample size n > 30
  • Sample mean = \overline{x}
  • Sample standard deviation = s
  • Population standard deviation = \sigma
  • Level of significance = \alpha

Then the confidence interval is obtained as

  • Case 1: Population standard deviation is known

\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}

  • Case 2: Population standard deviation is unknown.

\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}

For this case, we're given that:

  • Sample size n = 90 > 30
  • Sample mean = \overline{x} = 138
  • Sample standard deviation = s = 34
  • Level of significance = \alpha = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).

At this level of significance, the critical value of Z is: Z_{0.1/2} = ±1.645

Thus, we get:

CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]

Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C:  [130.10, 143.90]

Learn more about confidence interval for population mean from large samples here:

brainly.com/question/13770164

3 0
2 years ago
A sector with a central angle measure of 200 degrees has a radius of 9 cm. What is the area of the sector?
Sergio [31]

Answer:

\boxed{Area\ of\ sector = 141.4\ cm^2}

Step-by-step explanation:

Radius = r = 9 cm

Angle = θ = 200° = 3.5 radians

Now,

Area \ of \ sector = \frac{1}{2} r^2 \theta

Area = 1/2 (9)²(3.5)

Area = 1/2 (81)(3.5)

Area = 282.7 / 2

Area of sector = 141.4 cm²

6 0
3 years ago
Read 2 more answers
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