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professor190 [17]

2 years ago

6

Which rectangle is not an enlargement of rectangle A? ОА ОВ ОС OD

2 answers:

soldier1979 [14.2K]2 years ago

7 0

Answer: D

Step-by-step explanation:

Triangle D is not a enlargement of Triangle A because it is not the same type of number. 3 is an odd number, while 60 is an even number. The answer is D.

Hope this helps!

ExtremeBDS [4]2 years ago

7 0

Answer:

Hi guys! The answer is D

Step-by-step explanation:

I took it on edgen and got it corrected have a good,stay safe, and stay happy

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Can someone help me with this math homework please!

kobusy [5.1K]

Answer:

It's 2, 1, and y = 2x + 1.

Step-by-step explanation:

You can see the rise is 2 and the run is 1, making the slope = 2, and the y-intercept is 1 because that is where it crosses the y axis. Once you have the slope and y intercept, you can put it in a function, with the form being y=2x+1, the slope being the number before the x and the y-int value being after the x.

8 0

2 years ago

Read 2 more answers

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

ExtremeBDS [4]

The answer is in my attachment

6 0

3 years ago

I need to know what the answer is

wel

(x/2)-(5x/6)=1/9
x=-1/3

7 0

3 years ago

If the equation 3x-5y=-3, what is the value of y when x is 1

pantera1 [17]

The answer is: 6/5

Here is why:
3x - 5y = -3
3(1) - 5y = -3
3 - 5y = -3
3 - 5(6/5) = -3
-3 = -3

6 0

3 years ago