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erik [133]
3 years ago
5

I need help with this please have the answers for all the blank boxes

Mathematics
2 answers:
m_a_m_a [10]3 years ago
8 0
In order from left to right the answers are -6x 4x -42
Soloha48 [4]3 years ago
7 0

Answer:

Answers are -6x 4x -42?

Hopefully, I am right?!

:))

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Suppose that 16% of the population of the U.S. is left-handed. If a random sample of 170 people from the U.S. is chosen, approxi
Shalnov [3]

Answer:

0.6064 = 60.64% probability that fewer than 29 are left-handed.

Step-by-step explanation:

Binomial probability distribution

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

Can be approximated to a normal distribution, using the expected value and the standard deviation.

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)}

Normal probability distribution

Problems of normally distributed distributions 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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

16% of the population of the U.S. is left-handed.

This means that p = 0.16

Sample of 170 people

This means that n = 170

Mean and standard deviation:

\mu = E(X) = np = 170*0.16 = 27.2

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{170*0.16*0.84} = 4.78

Probability that fewer than 29 are left-handed.

Using continuity correction, this is P(X < 29 - 0.5) = P(X < 28.5), which is the pvalue of Z when X = 28.5. So

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

Z = \frac{28.5 - 27.2}{4.78}

Z = 0.27

Z = 0.27 has a pvalue of 0.6064

0.6064 = 60.64% probability that fewer than 29 are left-handed.

6 0
3 years ago
Which of the following best describes the end behavior of y=x^2 -bx +c as x approaches either positive or negative infinity?
Tcecarenko [31]
As x approaches either positive or negative infinity, y will approach positive infinity. 
This is because x^2 will grow large much quicker than b*x, so it has more importance. So when considering the limit, we can imagine that the equation looks like this:
y = x^2 
This equation will go to infinity for x approaching both negative and positive infinity.
7 0
3 years ago
Select the correct answer.
yan [13]

Answer:

the money spent by a buyer to purchase a product

6 0
3 years ago
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Solve |x-3|=7 plz help asap I will mark brainliest
Liula [17]

Answer:

x=10  or x=-4

Step-by-step explanation:

|x-3|=7

There are two solutions, one positive and one negative

x-3 = 7   and x-3 = -7

Add 3 to all sides

x-3+3 = 7+3           x-3+3 = -7+3

x = 10                    x = -4

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3 years ago
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PLS HELP!!!! Pointtssss!! The temperature during a very cold day is
grin007 [14]

The type of polynomial that would best model the data is a <em>cubic</em> polynomial. (Correct choice: D)

<h3>What kind of polynomial does fit best to a set of points?</h3>

In this question we must find a kind of polynomial whose form offers the <em>best</em> approximation to the <em>point</em> set, that is, the least polynomial whose mean square error is reasonable.

In a graphing tool we notice that the <em>least</em> polynomial must be a <em>cubic</em> polynomial, as there is no enough symmetry between (10, 9.37) and (14, 8.79), and the points (6, 3.88), (8, 6.48) and (10, 9.37) exhibits a <em>pseudo-linear</em> behavior.

The type of polynomial that would best model the data is a <em>cubic</em> polynomial. (Correct choice: D)

To learn more on cubic polynomials: brainly.com/question/21691794

#SPJ1

6 0
2 years ago
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