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Paladinen [302]

3 years ago

14

I believe some of the brainly admins are abusing their power because they keep deleting my questions and I am not violating guid

elines and I read them over and over

1 answer:

VladimirAG [237]3 years ago

6 0

I agree. Some people's questions keep getting deleted. Thanks you for bringing this up. Hopefully someone can contact the administrator.

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Which triangles (if any) can be shown to be congruent using a rotation?

-BARSIC- [3]

B...<span>Triangles </span>2 and 3<span> are </span>rotations<span> of each other. </span>Triangle<span> 1 requires a </span>reflection<span>.</span>

7 0

3 years ago

Read 2 more answers

Helppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp

kobusy [5.1K]

Answer:

2. -1/3x > 12

3. 5+x > 7

4.10-x < 30

5. 2+5x ≤ 3

6.6-2x ≥ 17

Step-by-step explanation:

7 0

3 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

3 years ago

Three students are playing a game using the spinner above. Each student draws three boxes on a piece of paper. The spinner is sp

harina [27]

Answer:

Probability

Step-by-step explanation:

In this game, the players rely entirely on chance to win, rather than skill. Therefore, this game would be particularly useful in helping students' understanding of probability.

(Spoken from someone who played this exact game 2 years ago to learn about probability.)

5 0

3 years ago

What number times itself equals 196?

leonid [27]

<span>number times itself equals 196
</span>
$= n * n = 196$

$14 * 14 = 196$

8 0

3 years ago

Read 2 more answers