Answer:
x = 25°
Step-by-step explanation:
2x-10°+2x-10°+100°=180°[sum of angles in a triangle ]
4x+80°=180°
4x=100°
x=25°
Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:
X = 590:


Z = 0.76
Z = 0.76 has a p-value of 0.7764.
X = 400:


Z = -0.89
Z = -0.89 has a p-value of 0.1867.
0.7764 - 0.1867 = 0.5897 = 58.97%.
58.97% of students would be expected to score between 400 and 590.
More can be learned about the normal distribution at brainly.com/question/27643290
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For a smoothing constant of 0.2
Time period – 1 2 3 4 5 6 7 8 9 10
Actual value – 46 55 39 42 63 54 55 61 52
Forecast – 58 55.6 55.48 52.18 50.15 52.72 52.97 53.38 54.90
Forecast error - -12 -.6 -16.48 – 10.12 12.85 1.28 2.03 7.62 -2.9
The mean square error is 84.12
The mean forecast for period 11 is 54.38
For a smoothing constant of 0.8
Time period – 1 2 3 4 5 6 7 8 9 10
Actual value – 46 55 39 42 63 54 55 61 52
Forecast – 58 48.40 53.68 41.94 41.99 58.80 54.96 54.99 59.80
Forecast error - -12 6.60 -14.68 0.06 21.01 -4.80 0.04 6.01 -7.80The mean square error is 107.17
The mean forecast for period 11 is 53.56
Based on the MSE, smoothing constant of .2 offers a better model since the mean forecast is much better compared to the 53.56 of the smoothing constant of 0.8.
Answer:
its one solution
Step-by-step explanation: