Please help me with this

Mathematics

1 answer:

zhannawk [14.2K]2 years ago

8 0

Answer: -6, -3, -2, -1, 1, 2, 3, 6

<u>Step-by-step explanation:</u>

NOTES: p is the last term, q is the coefficient of the first term

Determine all of the factors of p and all of the factors of q

Possible rational roots are all of the combinations of p/q for every factor

$\dfrac{p}{q}=\pm \dfrac{6}{1}=\pm \dfrac{1\times 6, 2\times 3}{1}=\pm\dfrac{1}{1}, \pm\dfrac{6}{1}, \pm\dfrac{2}{1}, \pm\dfrac{3}{1}\\\\\\\text{In order from smallest to largest:}\ \large\boxed{-6, -3, -2, -1, +1, +2, +3, +6}$

You might be interested in

F(x)=x²-3 tell whether the function is quadratic or not?

Misha Larkins [42]

Answer:

state whether this function is quadratic? why or why not?

f(x) = - 3(x - 2)(x + 3)(x + 5)

7 0

2 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>