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KonstantinChe [14]
3 years ago
8

Greatest common factor of 33 and 66

Mathematics
2 answers:
Paladinen [302]3 years ago
6 0

Gif:33 See image to see how I got it. By factorizing

Firdavs [7]3 years ago
4 0

the answer is 11 hope this helps

brainliest

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Solve the equation of isosceles triangle and find the x​
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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°

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Find the volume of a cube that has a side length of:<br> 6x²y5
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A college conducts a common test for all the students. For the Mathematics portion of this test, the scores are normally distrib
Jet001 [13]

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 \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • 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:

\mu = 502, \sigma = 115

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 = \frac{X - \mu}{\sigma}

Z = \frac{590 - 502}{115}

Z = 0.76

Z = 0.76 has a p-value of 0.7764.

X = 400:

Z = \frac{X - \mu}{\sigma}

Z = \frac{400 - 502}{115}

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

#SPJ1

6 0
1 year ago
The number of pizzas ordered on friday evenings between 5:30 and 6:30 at a pizza delivery location for the last 10 weeks is show
Ainat [17]
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.
7 0
3 years ago
How many solutions does this system of equations have?
azamat

Answer:

its one solution

Step-by-step explanation:

3 0
2 years ago
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