What is equal to 0/0 0 0 0 none of the above

Are these two claims equivalent, in conflict, or not comparable because they're talking about different things?9 a. "75 percent

blsea [12.9K]

Step-by-step explanation:

dnxjdjedjxkdkdkxkxmxmdmxmcmcmxmxm cxvfbfb cfbfbfbfbt xswzxvt

8 0

2 years ago

Find the missing number to make these fractions equal. 3/4 = ?/8

babymother [125]

The answer is 6, reaosn is because 6/8 is not simplified, so if we divide both sides by 2 (the numerator and the denominator), we can get a simplified fraction, which is 3/4. Steps: 6/8, 6 divided by 2/8 divided by 2, 3/4.

Hope this helped!

Nate

7 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

6. Write and solve a system of THREE Inequalities that models the situation below.

yuradex [85]

Answer: uh.. 3?

Step-by-step explanation:

lol

7 0

1 year ago

For the figures shown, HJKL ~ QRST, and the scale factor of HJKL to QRST is 9/7.

ch4aika [34]

Hey there! :)

JK ≈ RS

Scale Factor = 9/7

JK = 56

56 · 9 = 504 = 72
1 7 = 7

Your answer ⇒ B.72

Hope this helps :)

4 0

2 years ago