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OverLord2011 [107]
3 years ago
6

Multiply each polynomial (x - 7)(x - 4)

Mathematics
1 answer:
german3 years ago
8 0

Answer:2x+28 100% right

Step-by-step explanation:

You might be interested in
The Gcf of 6,14,18,and the Lcd of 5,9,13
mrs_skeptik [129]
Gcf of 6,14,18: 2
And you mean lcm?
6 0
3 years ago
24. Which measurement is equivalent to 10 g?<br> a. 1 dag<br> b. 1 dg<br> c. 1 cg<br> d. 1 hg
Liono4ka [1.6K]

Answer:

  a.  1 dag

Step-by-step explanation:

The list of SI prefixes is attached. Each has an abbreviation that can be used with the abbreviation for the SI unit of interest.

<h3>Application</h3>

A factor of 10 is signified by the prefix deka-, abbreviated "da". Then 10 grams is 1 dekagram, or 1 dag.

7 0
2 years ago
Explain linear function. Provide a real-world example of linear function, explain the rate of change. Include reference(s) in yo
GuDViN [60]

Answer:

See step-by-Step explanation

Step-by-step explanation:

An linear equation is an equation that shows the relationship of 2 numbers, and when one variable decreases at a constant rate, the other will also decrease by a constant rate. One example is when you drive at a constant rate at 50 miles per hour. The distance is y miles, and the time is x hours. The linear equation is y=50x.

8 0
3 years ago
From the graph of the function, determine the domain and the range.
harina [27]

Answer:

Domain(-4,4) and Range:(-3,0)

Step-by-step explanation:

Because the graph goes through - 4 and 4

3 0
2 years ago
. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

Problems of normally distributed samples can be 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.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

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

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
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