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Natalka [10]
3 years ago
5

{101, 2, 108, 104, 102, 107, 109, 102}

Mathematics
1 answer:
Bumek [7]3 years ago
4 0

Answer:

a should be your answer

Step-by-step explanation:

You might be interested in
Polygon MNOPQ is dilated by a scale factor of 0.8 with the origin as the center of dilation, resulting in the image M’N’O’P’Q’.
N76 [4]
Answer:
slope of M'N' = 1

Explanation:
First, we will need to get the coordinates of points M' and N':
We are given that the dilation factor (k) is 0.8
Therefore:
For point M':
x coordinate of M' = k * x coordinate of M 
x coordinate of M' = 0.8 * 2 = 1.6
y coordinate of M' = k * y coordinate of M
y coordinate of M' = 0.8 * 4 = 3.2
Therefore, coordinates of M' are (1.6 , 3.2)

For point N':
x coordinate of N' = k * x coordinate of N 
x coordinate of N' = 0.8 * 3 = 2.4
y coordinate of N' = k * y coordinate of N
y coordinate of N' = 0.8 * 5 = 4
Therefore, coordinates of M' are (2.4 , 4)

Then, we can get the slope of M'N':
slope = (y2-y1) / (x2-x1)
For M'N':
slope = (3.2-4) / (1.6-2.4)
slope = 1

Hope this helps :)
8 0
4 years ago
The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co
Alona [7]

Answer:

95% Confidence interval for the variance:

3.6511\leq \sigma^2\leq 34.5972

95% Confidence interval for the standard deviation:

1.9108\leq \sigma \leq 5.8819

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

s^2=2.89^2=8.3521

The confidence interval for the variance is:

\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128

Then, the confidence interval can be calculated as:

\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819

6 0
3 years ago
One man is three times as old as another. Fifteen years ago the first was six times as old as the second. Find their ages now.
alex41 [277]

Answer:

error

Step-by-step explanation:

4 0
3 years ago
*<br> Write the equation in standard form using integers.<br><br> y=-4/5x+3
svetoff [14.1K]

Answer:

4x + 5 y = 15

Step-by-step explanation:

6 0
3 years ago
6 times the sum of a number t and 5 is s
Crank
Umm simplified it would be 30t=s.  6*5t=s
6 0
3 years ago
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