This is dueeeeeee tomorow please help

Please help my exam is going on please and both question of math

omeli [17]

Step-by-step explanation:

11) PnQ=QnP

or, {1,2,3}n{1,2,3,4,5} = {1,2,3,4,5} n{1,2,3}

or, {1,2,3} = {1,2,3} .Ans

5 0

2 years ago

Which of the following statements about EnergyStar washing machines and dryers is false?

sineoko [7]

Which of the following statements about EnergyStar washing machines and dryers is false?
a.EnergyStar washers and dryers consume 30 to 40 percent less energy per load.b.EnergyStar top-loading washers use less water than front-loading washers.c.EnergyStar washers spin-dry laundry, which reduces dryer usage.d.EnergyStar washers and dryers have less tear and wear on laundry.

Answer is B

6 0

2 years ago

Explain how you can use mental math to compare 7 pounds to 120 ounces.

Eva8 [605]

One pound consists of 16oz, you have 120oz. If you divide 120 and 16 you get 7.5 lbs. so therefore you have more oz than lbs.

3 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

I don’t need to know how to solve it I’m going to ask my teacher how to tomorrow but for now I need help on these

Mars2501 [29]

1) slope=3 and y-int.=-5
2) slope=2 and y-int.=-6
3) slope=-6 and y-int.=1/2
4) slope=-7 and y-int.=5/2
5) slope=1/2 and y-int.=7
6) slope=3/4 and y-int.=8
7) slope=-2/3 and y-int.=-1/3
8) slope=-1/8 and y-int.=-3/8
9) slope=2/3 and y-int.=5
10) slope=-2/7 and y-int.=-1
11) slope=-3 and y-int.=6
12) slope=4 and y-int.=7
Hope this helps!

4 0

2 years ago