What is the resulting polynomial when 3x+7 is multiplied by 2x-6

Semenov [28]

So the answer is 13.5

Look at the attached picture⤴

Hope it will help u..

4 0

3 years ago

Read 2 more answers

There are 210210210 students in a twelfth grade high school class. 909090 of these students have at least one sister and 1051051

BlackZzzverrR [31]

Answer:

(1) The value of P (A) is 0.4286.

(2) The value of P (B) is 0.50.

(3) The value of P (A ∩ B) is 0.2143.

(4) The the value of P (B|A) is 0.50.

(5) The events A and B are independent.

Step-by-step explanation:

The events are defined as follows:

A = a student in the class has a sister

B = a student has a brother

The information provided is:

N = 210

n (A) = 90

n (B) = 105

n (A ∩ B) = 45

The probability of an event E is the ratio of the favorable number of outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

The conditional probability of an event X provided that another event Y has already occurred is:

$P(X|Y)=\frac{P(A\cap Y)}{P(Y)}$

If the events X and Y are independent then,

$P(X|Y)=P(X)$

(1)

Compute the probability of event A as follows:

$P(A)=\frac{n(A)}{N}\\\\=\frac{90}{210}\\\\=0.4286$

The value of P (A) is 0.4286.

(2)

Compute the probability of event B as follows:

$P(B)=\frac{n(B)}{N}\\\\=\frac{105}{210}\\\\=0.50$

The value of P (B) is 0.50.

(3)

Compute the probability of event A and B as follows:

$P(A\cap B)=\frac{n(A\cap B)}{N}\\\\=\frac{45}{210}\\\\=0.2143$

The value of P (A ∩ B) is 0.2143.

(4)

Compute the probability of B given A as follows:

$P(B|A)=\frac{P(A\cap B)}{P(A)}\\\\=\frac{0.2143}{0.4286}\\\\=0.50$

The the value of P (B|A) is 0.50.

(5)

The value of P (B|A) = 0.50 = P (B).

Thus, the events A and B are independent.

6 0

3 years ago

Click here What is the average rate of change for this exponential function for the interval from x=2 to x = 4? 15 10 -5 5 10 O

Triss [41]

Answer:

c. 6

Step-by-step explanation:

Given

See attachment for graph

Required

Average rate of change

This is calculated as:

$Rate = \frac{f(b) - f(a)}{b - a}$

Where

$(a,b) = (2,4)$

So, we have:

$Rate = \frac{f(4) - f(2)}{4 - 2}$

$Rate = \frac{f(4) - f(2)}{2}$

Using the attached graph, we have:

$f(2) = 4$

$f(4) = 16$

So, we have:

$Rate = \frac{16- 4}{2}$

$Rate = \frac{12}{2}$

$Rate = 6$

4 0

2 years ago

3/14 x 3/14 as a fraction

nataly862011 [7]

Answer:

((3 / 14) * 3) / 14 = 9 / 196

Step-by-step explanation:

Reduce 9/196 to lowest terms 9 196 is already in the simplest form. It can be written as 0.045918 in decimal form (rounded to 6 decimal places).

5 0

2 years ago

Philip Morris wishes to determine if there is a difference between the proportion of women and proportion of men who smoke cigar

uranmaximum [27]

Answer:

Step-by-step explanation:

This is a test of 2 population proportions. Let 1 and 2 be the subscript for the women and men who smoke respectively. The population proportion of women and men who smoke would be p1 and p2 respectively.

p1 - p2 = difference in the proportion of women and men who smoke.

The null hypothesis is

H0 : p1 = p2

p1 - p2 = 0

The alternative hypothesis is

Ha : p1 > p2

p1 - p2 > 0

it is a right tailed test

Sample proportion = x/n

Where

x represents number of success(number of complaints)

n represents number of samples

For women

x1 = 14

n1 = 110

p1 = 14/110 = 0.13

For men,

x2 = 7

n2 = 95

p2 = 7/95 = 0.07

The pooled proportion, pc is

pc = (x1 + x2)/(n1 + n2)

pc = (14 + 7)/(110 + 95) = 0.1

1 - pc = 1 - 0.1 = 0.9

z = (p1 - p2)/√pc(1 - pc)(1/n1 + 1/n2)

z = (0.13 - 0.07)/√(0.1)(0.9)(1/110 + 1/95)

z = 1.47

From the options given, the test statistic is

e. none of these is correct.

Since it is a right tailed test, we will look at the area to the right of z = 1.47. The p value would be

p value = 1 - 0.9292 = 0.078

Since alpha, 0.005 < than the p value, 0.078, then we would fail to reject the null hypothesis. Therefore, the data does not indicate that the proportion of women who smoke cigarettes is higher than the proportion of men who do at α=.005

8 0

2 years ago