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3 years ago

Given p(x) = x4 + x3 - 13x2 - 25x - 12

1 answer:

3 0

Answer + Explanation + Theory

When a number is divided by a number it results in a quotient and a remainder

E.g. 9 / 4 = 2 remainder 1

9 is the dividend
4 is the divisor
2 is the quotient
1 is the remainder

Same way when a polynomial is divided by a linear expression

E.g.

Ax^2 + bx + c / (x-b) = (x+a) + r

Which can be rearranged to

ax^2 + bx + c = (x+a)(x-b) + r

When x = - a or b, only the remainder is left since either (x+a)(x-b) is 0.

If x = - a or b is substituted into the polynomial and the value is 0 then there is no remainder,

This would suggest (x+a) or (x-b) are factors of the polynomial.

Now apply this logic to these questions

1. Let’s assume (x-4) is a factor, this would mean that when x=4 is substituted into the polynomial the answer would be 0.

This is the case, therefore the remainder is 0.

2. Having seen the logic above (applied using the remainder and factor theorem) the linear expression is a factor of the polynomial.

You might be interested in

6,500 acres of a 41,000 acre forest burned in a fire. what percentage of forest was damaged?

Tems11 [23]

Answer:

15.85%

Step-by-step explanation:

Given:

Acres damaged (burned) = 6,500 acres

Acres in total forest = 41,000 acres

Hence the amount of damaged forest as a fraction of the total forest,

= $\frac{6500}{41000}$ = $\frac{13}{82}$

expressed as percentage = $\frac{13}{82}$ x 100% = 15.85%

8 0

2 years ago

How do l do this I’m so confused

Anika [276]

Answer:

$967.6cm^{2} (1 d.p)$

Step-by-step explanation:

Total Surface Area of Cylinder = $2\pi rh+2\pi r^{2}$

Given from the question, r = 7 cm and h = 15cm

Lets Substitute r and h into the formula to find the Total Surface Area of the cylinder.

Total Surface Area of Cylinder = $2\pi (7)(15)+2\pi (7)^{2} \\=210\pi +98\pi \\=308\pi cm^{2} \\=967.6cm^{2} (1 d.p)$

7 0

2 years ago

The amount of pollutants that are found in waterways near large cities is normally distributed with mean 8.6 ppm and standard de

Setler79 [48]

We assume that question b is asking for the distribution of $\\ \overline{x}$ , that is, the distribution for the average amount of pollutants.

Answer:

a. The distribution of X is a normal distribution $\\ X \sim N(8.6, 1.3)$ .

b. The distribution for the average amount of pollutants is $\\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})$ .

c. $\\ P(z>-0.08) = 0.5319$ .

d. $\\ P(z>-0.47) = 0.6808$ .

e. We do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for $\\ \overline{X}$ is also normal because the sample was taken from a normal distribution.

f. $\\ IQR = 0.2868$ ppm. $\\ Q1 = 8.4566$ ppm and $\\ Q3 = 8.7434$ ppm.

Step-by-step explanation:

First, we have all this information from the question:

The random variable here, X, is the number of pollutants that are found in waterways near large cities.
This variable is normally distributed, with parameters:
$\\ \mu = 8.6$ ppm.
$\\ \sigma = 1.3$ ppm.
There is a sample of size, $\\ n = 38$ taken from this normal distribution.

a. What is the distribution of X?

The distribution of X is the normal (or Gaussian) distribution. X (uppercase) is the random variable, and follows a normal distribution with $\\ \mu = 8.6$ ppm and $\\ \sigma =1.3$ ppm or $\\ X \sim N(8.6, 1.3)$ .

b. What is the distribution of $\\ \overline{x}$ ?

The distribution for $\\ \overline{x}$ is $\\ N(\mu, \frac{\sigma}{\sqrt{n}})$ , i.e., the distribution for the sampling distribution of the means follows a normal distribution:

$\\ \overline{X} \sim N(8.6, \frac{1.3}{\sqrt{38}})$ .

c. What is the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants?

Notice that the question is asking for the random variable X (and not $\\ \overline{x}$ ). Then, we can use a standardized value or z-score so that we can consult the standard normal table.

$\\ z = \frac{x - \mu}{\sigma}$ [1]

x = 8.5 ppm and the question is about $\\ P(x>8.5)$ =?

Using [1]

$\\ z = \frac{8.5 - 8.6}{1.3}$

$\\ z = \frac{-0.1}{1.3}$

$\\ z = -0.07692 \approx -0.08$ (standard normal table has entries for two decimals places for z).

For $\\ z = -0.08$ , is $\\ P(z$ .

But, we are asked for $\\ P(z>-0.08) \approx P(x>8.5)$ .

$\\ P(z-0.08) = 1$

$\\ P(z>-0.08) = 1 - P(z$

$\\ P(z>-0.08) = 0.5319$

Thus, "the probability that one randomly selected city's waterway will have more than 8.5 ppm pollutants" is $\\ P(z>-0.08) = 0.5319$ .

d. For the 38 cities, find the probability that the average amount of pollutants is more than 8.5 ppm.

Or $\\ P(\overline{x} > 8.5)$ ppm?

This random variable follows a standardized random variable normally distributed, i.e. $\\ Z \sim N(0, 1)$ :

$\\ Z = \frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$ [2]

$\\ z = \frac{\overline{8.5} - 8.6}{\frac{1.3}{\sqrt{38}}}$

$\\ z = \frac{-0.1}{0.21088}$

$\\ z = \frac{-0.1}{0.21088} \approx -0.47420 \approx -0.47$

$\\ P(z$

Again, we are asked for $\\ P(z>-0.47)$ , then

$\\ P(z>-0.47) = 1 - P(z$

$\\ P(z>-0.47) = 1 - 0.3192$

$\\ P(z>-0.47) = 0.6808$

Then, the probability that the average amount of pollutants is more than 8.5 ppm for the 38 cities is $\\ P(z>-0.47) = 0.6808$ .

e. For part d), is the assumption that the distribution is normal necessary?

For this question, we do not need to assume that the distribution from we take the sample is normal. We already know that the distribution for X is normally distributed. Moreover, the distribution for $\\ \overline{X}$ is also normal because the sample was taken from a normal distribution. Additionally, the sample size is large enough to show a bell-shaped distribution.

f. Find the IQR for the average of 38 cities.

We must find the first quartile (25th percentile), and the third quartile (75th percentile). For $\\ P(z$ , $\\ z \approx -0.68$ , then, using [2]: