Write the following equation symbolically.

Use the rational zero thereom to list all possible zeroes of the function f(x)= 5x^8-2x^5-7

Scrat [10]

Possible rational zeros are:

$\pm1,\pm7,\pm\frac{1}{5},\pm\frac{7}{5}$

Step-by-step explanation:

We need to list all possible zeroes of the function $f(x)= 5x^8-2x^5-7$ using rational zero theorem.

The rational zero theorem.

rational zero theorem states that zeros can be found by dividing the factors of constant term called p by the factors of the leading co-efficient called q i.e

p/q

In our given function $f(x)= 5x^8-2x^5-7$

p = 7

q = 5

Factors of p (7) = ±1, ±7

Factors of q (5) = ±1, ±5

$\frac{p}{q} =\frac{\pm1,\pm7}{\pm1,\pm5}$

So, possible rational zeros are:

$\pm1,\pm7,\pm\frac{1}{5},\pm\frac{7}{5}$

Keywords: rational zero theorem, possible zeroes

Learn more about rational zero theorem at:

#learnwithBrainly

4 0

3 years ago

Two corporate baseball teams are scheduled to play a game together. They agree that if both teams attend or if neither team atte

sp2606 [1]

Answer:

2.99% probability that the cost will be paid by only one team

Step-by-step explanation:

The binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability that a team plays:

A team plays if it has at most 2 injured players out of 11.

11 players, so $n = 11$

Each player with a 5% probability of injury, so $p = 0.05$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{11,0}.(0.05)^{0}.(0.95)^{11} = 0.5688$

$P(X = 1) = C_{11,1}.(0.05)^{1}.(0.95)^{10} = 0.3293$

$P(X = 2) = C_{11,2}.(0.05)^{2}.(0.95)^{9} = 0.0867$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.5688 + 0.3293 + 0.0867 = 0.9848$

Each team has a 0.9848 probability of showing up to play.

What is the probability that the cost will be paid by only one team?

This happens if one team shows up and the other do not.

2 teams, so $n = 2$

Each team has a 0.9848 probability of showing up to play, so $p = 0.9848$ .

This probability is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{2,1}.(0.9848)^{1}.(0.0152)^{1} = 0.0299$

2.99% probability that the cost will be paid by only one team

7 0

4 years ago

The expression above can also be written in the form <br> So what is A =

nydimaria [60]

Answer:

what is the expressio

Step-by-step explanation:

3 0

2 years ago

Select the correct answer.

slavikrds [6]

Answer:

the answer is B

$3( \frac{7}{5} x + 4) - 2( \frac{ 3}{2} - \frac{5}{4} x)$

$(3 \times \frac{7}{5} x) + (3 \times 4) - 2( \frac{3}{2} - \frac{5}{4} x)$

$\frac{21}{5} x + 12 - 2( \frac{3}{2} - \frac{5}{4} x)$

$\frac{21}{5} x + 12 + ( - 2 \times \frac{3}{2} ) + ( - 2 \times ( - \frac{5}{4} x))$

$\frac{21}{5} x + 12 - 3 + \frac{5}{2} x$

$\frac{21}{5}x + \frac{5}{2} x + 12 - 3$

$\frac{67}{10}x - 9$

4 0

3 years ago

Eka mixes 7 6 7 pints of grape juice and 3 1 5 pints of cranberry juice to make punch for a party. She has 5 pints of punch left

Sliva [168]

Answer:

Eka served about 1,047 pints of punch at the party

Step-by-step explanation:

Total pints of grape and cranberry juices mixed = 767 + 315 = 1052 pints

Number of pints left over = 5 pints

Number of pints served = (Number of pints mixed) - (Number left over)

Number of pints served = 1052 - 5 = 1,047 pints

Eka served about 1,047 pints of punch at the party

3 0

3 years ago