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3 years ago

Question 7 of 10

Mathematics

1 answer:

jonny [76]3 years ago

5 0

The leading coefficient determines the shape of the graphs, such as how

the characteristic of a function are directed.

Correct response:

The graphs that represents functions that have a negative leading

coefficient are;

Graph A, Graph B, and Graph C

<h3>Methods by which the correct options are selected</h3>

Graph A:

The function in graph A is x³

When the coefficient of x³ is negative, the value of the function rises as x decreases from 0 to -∞, and decreases as x increases from 0 to ∞

Therefore;

The leading coefficient of Graph A is negative.

Graph B:

The value of the graph decreases as the magnitude of x increases

therefore, the graph is similar to a quadratic function, such that the leading

coefficient is negative, which inverts the function to give increasing output

with negative value as the value of x increases.

Therefore;

The leading coefficient of the quadratic function in Graph B is negative.

Graph C:

The given function in graph C is a linear function having a negative slope,

therefore;

The leading coefficient of x in the function in Graph C is negative.

Graph D:

The function of the graph in Graph D, that have y values that increases

exponentially as x increases is a quadratic function.

Given that y increases as the value of x increases, the leading coefficient

(coefficient of x²) is positive.

Therefore;

The graphs that represents functions that have negative leading coefficient are; Graph A, Graph B, and Graph C.

Learn more about the factors that depend on leading coefficient of a function here:

brainly.com/question/12209936

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$=\sqrt{9}\cdot \sqrt{a^3}$

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Rewrite. Note that 50 = 25(2):

$=\sqrt{25}\cdot \sqrt{2}\cdot \sqrt{b^4}$

Simplify. We can rewrite the factor as:

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The square and square root will cancel out. Thus:

$=5\sqrt{2}b^2$

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3 years ago

Read 2 more answers

A company that produces fine crystal knows from experience that 13% of its goblets have cosmetic flaws and must be classified as

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Answer:

(a) The probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b) The probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

Step-by-step explanation:

Let X = number of seconds in the batch.

The probability of the random variable X is, p = 0.31.

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}p^{x}(1-p)^{n-x};\ x=0,1,2,3...$

(a)

Compute the probability that only one goblet is a second among six randomly selected goblets as follows:

$P(X=1)={6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=6\times 0.13\times 0.4984\\=0.3888$

Thus, the probability that only one goblet is a second among six randomly selected goblets is 0.3888.

(b)

Compute the probability that at least two goblet is a second among six randomly selected goblets as follows:

P (X ≥ 2) = 1 - P (X < 2)

$=1-{6\choose 0}0.13^{0}(1-0.13)^{6-0}-{6\choose 1}0.13^{1}(1-0.13)^{6-1}\\=1-0.4336+0.3888\\=0.1776$

Thus, the probability that at least two goblet is a second among six randomly selected goblets is 0.1776.

(c)

If goblets are examined one by one then to find four that are not seconds we need to select either 4 goblets that are not seconds or 5 goblets including only 1 second.

P (4 not seconds) = P (X = 0; n = 4) + P (X = 1; n = 5)

$={4\choose 0}0.13^{0}(1-0.13)^{4-0}+{5\choose 1}0.13^{1}(1-0.13)^{5-1}\\=0.5729+0.3724\\=0.9453$

Thus, the probability that at most five must be selected to find four that are not seconds is 0.9453.

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