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uranmaximum [27]

2 years ago

12

FLIP OR NOT TO FLIP? A. -4x + 2 < -2 B. - 9>-8 -7.

1 answer:

vivado [14]2 years ago

5 0

Answer:

the answer is not flip because when u calculate it

You might be interested in

Are the irrational numbers closed under addition?

Arte-miy333 [17]

Answer:

Irrational numbers are not closed under addition.

Step-by-step explanation:

Irrational numbers are the numbers that cannot be expressed in the form of a fraction $\frac{x}{y}$ . In other words we can say that irrational number,s decimal expantion does not cease to end.

The closure property of addition in irrational numbers say that sum of two irrational number is always a rational number, But this is not true. It is not necessary that the sum is always irrational some time it may be rational.

This can be understood with the help of an example:

let (2+√2) and (-√2) be two irrational number. Their sum is (2+√2)+(-√2) = 2, which is clearly a rational number.

Hence, irrational numbers are not closed under addition.

4 0

2 years ago

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

(c) 0.0039

(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

2 years ago

Question: Carbon-11 has a half-life of about 20 minutes. If we start with 200 grams, how

irakobra [83]

Answer:

C). 0.39 grams

Step-by-step explanation:

First find how many half-lifes are in 3 hours:

3 hours ÷ 20 minutes = 9 half-lifes

Find the total left:

0.5^9*200 ≈ 0.39

3 0

2 years ago

Is (2, 4) a solution to the system<br> y = 2x<br> x + y = 6

posledela

Answer:

yes you plug the numbers in

5 0

2 years ago

Giving fr$e points your welcome! have a good rest of your week

victus00 [196]

Step-by-step explanation:

Thank you buddy

have a great day ahead

6 0

1 year ago

Read 2 more answers