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3 years ago

Find the measures of the numbered angles for each parallelogram. (See attachment)

Mathematics

2 answers:

Bogdan [553]3 years ago

7 0

Since the figure is a parallelogram, angle 3 is an alternate interior angle with the 50 deg. angle. Hence, m<3 = 50 deg. Angle 2 is supplementary with 82 deg. angle, therefore m<2 = 180 deg - 82 deg = 98 deg. Using angles 2 and 3, angle 1 can be calculated since the total interior angle of a triangle is 180 deg. m<1 = 180 - 98 - 50 = 32 deg.
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Alika [10]3 years ago

4 0

<2 = 180 - 82 = 98

<3 = 50

<1 = 180- ( 98 + 50)
<1 = 180 -148
<1 = 32

answer:
<1 = 32
<2 = 98
<3 = 50

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Additionally, to find the proportion of students who scored an A, the normal distribution is used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

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And p is the probability of X happening.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Proportion of students that scored an A:

Scores have a mean of 79 and a standard deviation of 11.3, which means that $\mu = 79, \sigma = 11.3$

Scores of 90 or higher are graded an A, which means that the proportion is 1 subtracted by the p-value of Z when X = 90, so:

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

$Z = \frac{90 - 79}{11.3}$

$Z = 0.97$

$Z = 0.97$ has a p-value of 0.8340.

1 - 0.8340 = 0.166

The proportion of students that scored an A is 0.166.

Probability that 6 students or more will score an "A" on the final exam:

Binomial distribution.

22 students, which means that $n = 22$

The proportion of students that scored an A is 0.166, which means that $p = 0.166$

The probability is:

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{22,0}.(0.166)^{0}.(0.834)^{22} = 0.0184$

$P(X = 1) = C_{22,1}.(0.166)^{1}.(0.834)^{21} = 0.0807$

$P(X = 2) = C_{22,2}.(0.166)^{2}.(0.834)^{20} = 0.1687$

$P(X = 3) = C_{22,3}.(0.166)^{3}.(0.834)^{19} = 0.2239$

$P(X = 4) = C_{22,4}.(0.166)^{4}.(0.834)^{18} = 0.2117$

$P(X = 5) = C_{22,5}.(0.166)^{5}.(0.834)^{17} = 0.1517$

Then

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$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8551 = 0.1449$

Thus

0.1449 = 14.49% probability that 6 students or more will score an "A" on the final exam.

For a problem that used the normal distribution, you can check brainly.com/question/15181104, and for a problem that used the binomial distribution, you can check brainly.com/question/15557838

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3 years ago