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valentina_108 [34]
3 years ago
10

Use the Rational Root Theorem to list all possible rational roots of the polynomial equation x3 – x2 – x – 3 = 0. Do not find th

e actual roots. (1 point)
–3, –1, 1, 3
1, 3
–33
no roots
Mathematics
2 answers:
almond37 [142]3 years ago
7 0
Hello,
<span>List all possible rational roots of the polynomial equation x3 – x2 – x – 3 = 0

Answer -1,1,-3,3
but none is a root.
</span>
Sauron [17]3 years ago
3 0

The Cubic Polynomial  given here is

x³ - x² - x -3 =0

Rational root theorem :

Consider any polynomial

→ Ax^n +A_{1}x^{n-1}+A_{2}x^{n-2}+.................+A_{n}=0

So, the factors of this polynomial is those number which divides \frac{A_{n}}{A}.Which are \pm1, \pm\frac{A_{n}}{A}

Applying the same method the number which divides (-3) are,-3,+3,-1,+1. So By rational root theorem all the factors of this cubic polynomial is \pm1,\pm3.

Option 1. -3,-1,1,3 is correct option.



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Yesterday I ran 5 miles. Today, I ran 3.7 Miles.
Stella [2.4K]

Answer:

percent decrease

26%

Step-by-step explanation:

Yesterday you ran 5 miles

Today you ran 3.7 miles

The amount went down, so the percent decreased

The percent decrease = (original - new)/original * 100 %

                                      = (5-3.7)/5 * 100%

                                      = 1.3/5 * 100 %

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5 0
3 years ago
Read 2 more answers
An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0
zmey [24]

Answer:

96.33% probability that everyone who appears for the flight will get a seat

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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 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.

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

In this problem, we have that:

n = 105, p = 0.9

So

\mu = E(X) = np = 105*0.9 = 94.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.9*0.1} = 3.07

What is the probability that everyone who appears for the flight will get a seat

100 or less people appearing to the flight, which is the pvalue of Z when X = 100. So

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

Z = \frac{100 - 94.5}{3.07}

Z = 1.79

Z = 1.79 has a pvalue of 0.9633

96.33% probability that everyone who appears for the flight will get a seat

7 0
3 years ago
Suppose an airline policy states that all baggage must be box shaped with a sum of​ length, width, and height not exceeding 114
NISA [10]

Answer:

Step-by-step explanation:

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Substituting 114 - 2s for h in the volume formula, we obtain:

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dV

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ds

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Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in

and the volume is V = s^2(h) = 46,298.25 in^3

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3 years ago
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