Ask question

Ask question

All categories

AleksandrR [38]

3 years ago

11

I need some help...

1 answer:

bixtya [17]3 years ago

7 0

-------------------------------------
Question
-------------------------------------
17x³ + (3x + 8x³<span>)

-------------------------------------
Remove the bracket
</span>-------------------------------------
17x³ + 3x + 8x³

-------------------------------------
Combine like terms
-------------------------------------
25x³ + 3x

-------------------------------------
Take out common factor
-------------------------------------
x(25x² + 3)

-------------------------------------
Answer: x(25x² + 3)
-------------------------------------

You might be interested in

I spent 80% of my money and had $12 left. How much money did I have to start ?

xeze [42]

If you spent 80% that means you have 20% left, we'll use B as variable

20% x B = 12
0.20/0.20 x B = 12/0.20
B = 60

6 0

4 years ago

What is an angle that measures 1/2 turn

Ad libitum [116K]

An angle that measures 1/2 turn is 180 degree angle

4 0

3 years ago

Which of the graphs in Fig. Q25.12 best illustrates the current I in a real resistor as a function of the potential difference V

AveGali [126]

What is the question

8 0

3 years ago

Math help j: pls Heh

BabaBlast [244]

Answer:

7/10

Step-by-step explanation:

0.7+0.07+0.007=0.777

fraction=777/1000

simplest form: 777/1000=7/10

srry if im wrong

hope this helps :3

8 0

3 years ago

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

3 years ago