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3 years ago

Combine the like terms to create an equivalent expression: 5n+6+(-7n)

Mathematics

2 answers:

adelina 88 [10]3 years ago

8 0

Answer:

-2n + 6

Step-by-step explanation:

To combine this expression, we simply put the like terms, the coefficient with the same variables, and constants together.

5n + 6 + (-7n)

So we can reorganize this:

(5n + -7n) + 6

==> (-2n) + 6

==> -2n + 6

So that is the equivalent expression with the combination of like terms.

Cheers.

prohojiy [21]3 years ago

7 0

Answer:

-2n +6

Step-by-step explanation:

The expression we are given is:

5n + 6 + (-7n)

We want to combine like terms. First, let's analyze each term.

5n ⇒ has a variable, "n"

5 ⇒ does not have a variable

-7n ⇒ also has a variable, "n"

The like terms in this case are 5n and -7n, since they both include a variable. Let's combine the like terms.

5n + 6 + (-7n)

(5n - 7n) +6

Subtract 7n from 5n.

(5-7) *n +6

(-2)*n +6

(-2n)+6

-2n + 6

There are no more like terms, so this is simplified as much as possible.

The expression 5n+6+(-7n) after combining the like terms is -2n+6.

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The sum of three consecutive even numbers is 48. What is the smallest of these numbers?

SVETLANKA909090 [29]

Answer: The numbers are 14, 16 and 18.

Step-by-step explanation: First thing to note is that consecutive even numbers have a difference of 2 units between every two terms. That is, if the first number is a, the next would be a + 2, and the next would be a + 4, and so on.

Therefore the three consecutive even numbers shall be a, a + 2, and a + 4.

Our equation now becomes;

a + a + 2 + a + 4 = 48

3a + 6 = 48

Subtract 6 from both sides of the equation

3a = 42

Divide both sides of the equation by 3

a = 14

Therefore, the numbers are 14, 16 and 18.

4 0

3 years ago

Read 2 more answers

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

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

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

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

By the Central Limit Theorem

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

$Z = \frac{20.4 - 18.6}{0.6768}$

$Z = 2.66$

$Z = 2.66$ has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

4 0

3 years ago

What is the equation of the line that passes through (4, -1)

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

answer: y = -5 + 19

We can use the point-slope formula to find an equation to solve this problem. The point-slope formula states: (y−y1)=m(x−x1)

Where m is the slope and (x1y1) is a point the line passes through.

Susbtituting the slope and values from the point from the problem gives:

(y−−1)=−5(x−4)

(y+1)=−5(x−4)

We can also solve this for the slope-intercept form. The slope-intercept form of a linear equation is: y=mx+b

Where m is the slope and b is the y-intercept value.

Substitute the slope from the problem for m and the values of the point from the problem for x and y and solve for b:

−1=(−5⋅4)+b

−1=−20+b

20−1=20−20+b

19=0+b

19=b

We can substitute for m and b in the formula to find the equation:

y=−5x+19

6 0

2 years ago

Read 2 more answers

A pathologist knows that 14.9% of all deaths are attributable to myocardial infarctions (a type of heart disease). (Either you h

garik1379 [7]

Answer:

a) The mean number is 745 and the standard deviation is 25.18.

b) 896 deaths is a significantly high number, which means that there is cause for concern.

Step-by-step explanation:

For each death, there are only two possible outcomes. Either it is attributable to myocardial infarctions, or it is not. The probability of a death being attributable to myocardial infarctions is independent of other deaths. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

A value X is significantly high is:

$X > E(X) + 2\sqrt{V(X)}$

In this problem:

$p = 0.149, n = 5000$

a. Find the mean and standard deviation for the number of such deaths that will occur in typical region with 5000 deaths.

$E(X) = np = 5000*0.149 = 745$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{5000*0.149*0.851} = 25.18$

The mean number is 745 and the standard deviation is 25.18.

b. In one region, 5000 death certificates are examined, and it is found that 896 deaths were attributable to myocardial infarction. Is there cause for concern? Why or why not?

Is 896 a significantly high number?

$E(X) + 2\sqrt{V(X)} = 745 + 2*25.18 = 795.36$

895 > 795.36

896 deaths is a significantly high number, which means that there is cause for concern.

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3 years ago

After getting a job at the ice cream parlor, Mitzi put $20 in

RideAnS [48]

Answer: The answer is C

Step-by-step explanation:

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3 years ago