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3 years ago

I need help on the Math HW. I don’t get how to get the answers.

Mathematics

1 answer:

shepuryov [24]3 years ago

4 0

Answer:

A= 34

B= 112

C= 34

Step-by-step explanation:

A and C are vertical angles so would have the same angle degree. The 112 and B are same side interior angles so have the same degree

A and 112 are opposite interior angles so they have to add up to 180.

So 180-112=68

then 68 divided by 2 to get both A and C

hope that made sense

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B= -2.75

A= -0.25

C= 2.50

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Solve the compound inequality.

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Answer:

Step-by-step explanation:

9 - 4x > = 5 or 4(-1 + x) - 6 > = 2

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4 years ago

15 - 3k = 10(k-5) Can someone please answer this

solmaris [256]

Answer:

k = 5

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS
Equality Properties

Step-by-step explanation:

Step 1: Define equation

15 - 3k = 10(k - 5)

Step 2: Solve for k

Distribute 10: 15 - 3k = 10k - 50
Add 3k to both sides: 15 = 13k - 50
Add 50 to both sides: 65 = 13k
Divide 13 on both sides: 5 = k
Rewrite: k = 5

Step 3: Check

Plug in k into the original equation to verify it's a solution.

Substitute in k: 15 - 3(5) = 10(5 - 5)
Subtract: 15 - 3(5) = 10(0)
Multiply: 15 - 15 = 0
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Here we see that 0 does indeed equal 0.

∴ k = 5 is a solution of the equation.

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3 years ago

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A professor knows that her statistics students' final exam scores have a mean of 79 and a standard deviation of 11.3. In his cla

OverLord2011 [107]

For each student, there are only two possible outcomes. Either they score an A, or they do not. The probability of a student scoring an A is independent of any other student, which means that the binomial probability distribution is used to solve this question.

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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Normal Probability Distribution

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

In a set with mean and standard deviation , the z-score of a measure X is given by:

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

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

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

Then

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

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

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

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

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

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

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

Then

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0184 + 0.0807 + 0.1687 + 0.2239 + 0.2117 + 0.1517 = 0.8551$

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

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

3 0

3 years ago