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2 years ago

Can anyone share the process please?

Mathematics

1 answer:

gtnhenbr [62]2 years ago

4 0

Answer:

3x³ + x² + x + 1

= 3x³ - 3x² + 4x² - 4x + 5x - 5 + 6

= 3x²(x - 1) + 4x(x - 1) + 5(x - 1) + 6

= (x - 1)(3x² + 4x + 5) + 6

but 3x³ + x² + x + 1 = (x - 1).Q(x) + R

=> Q(x) = 3x² + 4x + 5

R = 6

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De║bc, ad=2cm and bd=3cm, then area(Δabc):area(Δade) is equal to

xxTIMURxx [149]

Answer:

B.2:3

Step-by-step explanation:

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5 0

2 years ago

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

A = a student attended public high school

B = a student attended private high school

C = a student was home schooled

D = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0

2 years ago

"the quotient of a number and 4"

ladessa [460]

Answer:

a number divided by 4

Step-by-step explanation:

a number = n (it's a variable used in equations.)

so n over 4, or n divided by 4

3 0

3 years ago

Which of the following in an identity?

sammy [17]

By "which is an identity" they just mean "which trigonometric equation is true?"

What you have to do is take one of these and sort it out to an identity you know is true, or...

*FYI: You can always test identites like this:
Use the short angle of a 3-4-5 triangle, which would have these trig ratios:
sinx = 3/5 cscx = 5/3
cosx = 4/5 secx = 5/4
tanx = 4/3 cotx = 3/4
Then just plug them in and see if it works. If it doesn't, it can't be an identity!

Let's start with c, just because it seems obvious.
The Pythagorean identity states that sin²x + cos²x = 1, so this same statement with a minus is obviously not true.

Next would be d. csc²x + cot²x = 1 is not true because of a similar Pythagorean identity 1 + cot²x = csc²x. (if you need help remembering these identites, do yourslef a favor and search up the Magic Hexagon.)

Next is b. Here we have (cscx + cotx)² = 1. Let's take the square root of each side...cscx + cotx = 1. Now you should be able to see why this can't work as a Pythagorean Identity. There's always that test we can do for verification...5/3 + 3/4 ≠ 1, nor is (5/3 + 3/4)².

By process of elimination, a must be true. You can test w/ our example ratios:
sin²xsec²x+1 = tan²xcsc²x
(3/5)²(5/4)²+1 = (4/5)²(5/3)²
(9/25)(25/16)+1 = (16/25)(25/9)
(225/400)+1 = (400/225)
(9/16)+1 = (16/9)
(81/144)+1 = (256/144)
(81/144)+(144/144) = (256/144)
(256/144) = (256/144)

5 0

2 years ago

Find any points of discontinuity for the rational function.What are the points of discontinuity? Are they all removable?

UkoKoshka [18]

Hello,
Please, see the attached file.
Thanks.

7 0

3 years ago

Can anyone share the process please? ​

Can anyone share the process please?