Answer:
The answer to the question is;
The number of papers expected to be handed in before receiving each possible grade at least once is 14.93.
Step-by-step explanation:
To solve the question , we note that it is a geometric distribution question which have equal probabilities and therefore is a form of Binomial distribution with Bernoulli trials, where we are conducting the trials till we have r successes
Since we have r = 6, we will have to find the expected value of the number of trials till the nth paper handed in receives a previously awarded grade.
We therefore have,
The Probability that out of six papers turned 5 are different scores is given by
P(Y=5) = p'= q⁵p = (1-p)⁵p = 3125/46656
Therefore p' = the probability of receiving different grades once then the expected value is given by
E(X) = 1/p' = 46656/3125 = 14.93.
Answer:
64
Step-by-step explanation:
16 x 4 = 64
The difference between (5.29 times 10 superscript 11 baseline) minus (3.86 times 10 superscript 11 baseline) is 1. 43 × 10^11
<h3>How to determine the notation</h3>
Given the expression
(5. 29 × 10^11) - (3. 86 × 10 ^11)
First, find the common factor
10^11 ( 5. 29 - 3. 86)
Then substract the values within the bracket
10^11 (1. 43)
Multiply with the factor, we have
⇒1. 43 × 10^11
Thus, the difference between (5.29 times 10 superscript 11 baseline) minus (3.86 times 10 superscript 11 baseline) is 1. 43 × 10^11
Learn more about index notation here:
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2x + 8y = 5
2/ 24x - 4y = -15
2x + 8y = 5
48x - 8x = -30
+
--------------------------
50x = -25
x = -1/2
2 * -1/2 + 8y = 5
-1 + 8y = 5
8y = 6
y = 6/8
y = 3/4
x = -1/2 y = 3/4
If 1 in.= 2.54 cm, Then !9in. multiplied by 2.54 cm would be b. 48.26 cm
