All categories

3 years ago

Help plz ,I’m having some difficulties

Mathematics

2 answers:

kobusy [5.1K]3 years ago

6 0

Step-by-step explanation:

| x - 2 | < 4

Start at x=2. Go 4 units in either direction: x=-2 and x=6. Draw empty circles around each. Since it's less than, the line goes between the circles, connecting them. The result should look like this:

desmos.com/calculator/k2bujixjtx

| x + 2 | > 4

Start at x=-2. Go 4 units in either direction: x=-6 and x=2. Draw empty circles around each. Since it's greater than, the line goes outside of the circles. The result should look like this:

desmos.com/calculator/sk1cxob0tg

Murljashka [212]3 years ago

3 0

Answer:

Step-by-step explanation:

|x-2|<4

-4<x+2<4

subtract 2 from each inequality

-4-2<x+2-2<4-2

-6<x<2

|x+2|>4

x+2<-4 and x+2>4

or x+2-2<-4-2 and x+2-2>4-2

or x<-6 and x>2

You might be interested in

Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 1

Lesechka [4]

Answer:

(a) 0.28347

(b) 0.36909

(d) 0.9806

Step-by-step explanation:

Given information:

n=12

p = 20% = 0.2

q = 1-p = 1-0.2 = 0.8

Binomial formula:

$P(x=r)=^nC_rp^rq^{n-r}$

(a) Exactly two will be drunken drivers.

$P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}$

$P(x=2)=66(0.2)^{2}(0.8)^{10}$

$P(x=2)=\approx 0.28347$

Therefore, the probability that exactly two will be drunken drivers is 0.28347.

(b)Three or four will be drunken drivers.

$P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)$

$P(x=3\text{ or }x=4)=P(x=3)+P(x=4)$

Using binomial we get

$P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}$

$P(x=3\text{ or }x=4)=0.236223+0.132876$

$P(x=3\text{ or }x=4)\approx 0.369099$

Therefore, the probability that three or four will be drunken drivers is 0.3691.

(c)

At least 7 will be drunken drivers.

$P(x\geq 7)=1-P(x$

$P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)]$

$P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155]$

$P(x\leq 7)=1-[0.9961]$

$P(x\leq 7)=0.0039$

Therefore, the probability of at least 7 will be drunken drivers is 0.0039.

(d) At most 5 will be drunken drivers.

$P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)$

$P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315$

$P(x\leq 5)=0.9806$

Therefore, the probability of at most 5 will be drunken drivers is 0.9806.

5 0

3 years ago

You have 9 math coins and each coin is worth -25 points. How many points is that all together?

lesya [120]

The answer is -225.
9 times -25 is -225.
9*-25=-225

6 0

3 years ago

Read 2 more answers

Raquel can type 63 words every minnute.Rick type then Raquel in 135 minute?Circle the letter of the correct answer.A 1350 B 4599

lozanna [386]

Answer: 1350

Step-by-step explanation:

Here is the correct question.

Raquel can type an average of 63 words per minute. Rick can type 73 words per minute. how many more words can Rick type than Raquel in 135 minutes? Jared chose B as the correct answer. How did he get that answer? Jared said the answer is 4599. How did he get that answer?

Rick's word per minute= 73

Raquel's word per minute= 63

Ricks word in 135minute= 73×135 = 9855

Raquel's word in 135 minutes=63×135 = 8505

=9855 - 8505

= 1350

Rick can type 1350 more words than Raquel in 135 minutes.

Jared's answer is wrong. He got the answer by multiplying 63 by 73 which gives 4599.

8 0

3 years ago

Read 2 more answers

How would I do 4x+45=3x+61