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professor190 [17]

2 years ago

12

Write down the complement 17,30

1 answer:

rodikova [14]2 years ago

8 0

Be QuItE pEaSaNt!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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Point (a,a) lies on the graph of the equation 3y = 2x -4 find the value of a

Igoryamba

Substituting (x, y) with (a, a), since we have:

3a=2a-4, which means that 3a-2a=-4, that is a=-4.

So, the point is (-4, -4).

Remark: a point which lies in the graph of an equation is a solution of that equation. So x=a, y=a satisfies the equation <span> 3y = 2x -4, which makes our substitution legitimate.

Answer: </span>(-4, -4).

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3 years ago

Find the least length of a rope which can be cut into whole number of pieces of lengths 45cm, 75cm and 81 cm

choli [55]

The least is 45 bc it is less than the other 2

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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

Math, help plsssssssssss

Bingel [31]

Answer:

x-intercept= (2, 0)

y-intercept= (0,-6)

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2 years ago

Identify the median and the lower quartile for this box plot 16,16.5,15.5,

MArishka [77]

The median should be 16

8 0

2 years ago

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