The required probability is 
<u>Solution:</u>
Given, a shipment of 11 printers contains 2 that are defective.
We have to find the probability that a sample of size 2, drawn from the 11, will not contain a defective printer.
Now, we know that, 
Probability for first draw to be non-defective 
(total printers = 11; total defective printers = 2)
Probability for second draw to be non defective 
(printers after first slot = 10; total defective printers = 2)
Then, total probability 
<span>Let a_0 = 100, the first payment. Every subsequent payment is the prior payment, times 1.1. In order to represent that, let a_n be the term in question. The term before it is a_n-1. So a_n = 1.1 * a_n-1. This means that a_19 = 1.1*a_18, a_18 = 1.1*a_17, etc. To find the sum of your first 20 payments, this sum is equal to a_0+a_1+a_2+...+a_19. a_1 = 1.1*a_0, so a_2 = 1.1*(1.1*a_0) = (1.1)^2 * a_0, a_3 = 1.1*a_2 = (1.1)^3*a_3, and so on. So the sum can be reduced to S = a_0 * (1+ 1.1 + 1.1^2 + 1.1^3 + ... + 1.1^19) which is approximately $5727.50</span>
Answer:
76.2048 Kilograms
Step-by-step explanation:
You would multiply 168 by 0.4536
168 x 0.4536 = 76.2048
Answer:
.A -42+(-17)
Step-by-step explanation:
Answer:
<em><u>1=</u></em><em><u> </u></em> 237
<em><u>2=</u></em> 200
<em><u>3</u></em><em>= </em>49
<em><u>4=</u></em> 50
<em>I'm </em><em>99% </em><em>sure these are right.</em>
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<em> </em><u><em>please let me know if </em></u><u>any are wrong.</u>
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Have a good day<3</h2><h2>
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