6:20=9:30=3:10=18:60=16:160/3=12:40
Step-by-step explanation:
The question requires you to find value of x in the ratios given;
Start with the first pair
6:20 = 9:x
6/20 =9/x
6x=20*9
6x=180
x=180/6=30 ⇒⇒ 6:20 = 9:30
The second pair after replacing the value of x=30 will be
9:30 = x:10
9/30=x/10
90=30x
90/30 =x
3=x⇒⇒ 9:30 = 3:10
The third pair after replacing value of x=3 will be
3:10 =x:60
3/10 =x/60
180=10x
180/10 =x
18=x ⇒⇒ 3:10 = 18:60
The fourth pair after replacing value of x=18 will be;
18:60 = 16:x
18/60 = 16/x
18x=16*60
x= (16*60)/18 =160/3
x= 160/3 ⇒⇒⇒ 18:60 = 16: 160/3
The firth pair after replacing value of x=160/3 will be;
16: 160/3 =12:x
16x= 160/3 *12
16x = 160 * 4
x= (160 *4 )/16
x=40
⇒⇒ 16: 160/3 = 12: 40
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Ratios and proportions :brainly.com/question/9512748
Keywords : ratio, value of x, proportion
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Circumference~(pi)(diameter)~28.26
Area~ (pi)(radius)squared~63.59
I believe these are right
x= 5 x z= 3 = 15 (4y - 2z) = 60 - -30 = 30 so its C
The answer is D. -15 because I am assuming that the vertical lines mean brackets, and we have to work out the brackets first (according to BIDMAS). Doing so, -6 + 2 = -4 and -4 + -11 = -15.
Answer:
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.
Step-by-step explanation:
We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.
The probability of k online retail orders that turn out to be fraudulent in the sample is:

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:
![P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483](https://tex.z-dn.net/?f=P%28x%5Cgeq2%29%3D1-%5BP%28x%3D0%29%2BP%28x%3D1%29%5D%5C%5C%5C%5C%5C%5CP%28x%3D0%29%3D%5Cdbinom%7B20%7D%7B0%7D%5Ccdot0.08%5E%7B0%7D%5Ccdot0.92%5E%7B20%7D%3D1%5Ccdot1%5Ccdot0.189%3D0.189%5C%5C%5C%5C%5C%5CP%28x%3D1%29%3D%5Cdbinom%7B20%7D%7B1%7D%5Ccdot0.08%5E%7B1%7D%5Ccdot0.92%5E%7B19%7D%3D20%5Ccdot0.08%5Ccdot0.205%3D0.328%5C%5C%5C%5C%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-%5B0.189%2B0.328%5D%5C%5C%5C%5CP%28x%5Cgeq2%29%3D1-0.517%3D0.483)
The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.