Answer:
Step-by-step explanation:
From the given information; Let's assume that R should represent the set of all possible outcomes generated from a bit string of length 10 .
So; as each place is fitted with either 0 or 1
Similarly; the event E signifies the randomly generated bit string of length 10 does not contain a 0
Now;
if a 0 bit and a 1 bit are equally likely
The probability that a randomly generated bit string of length 10 does not contain a 0 if bits are independent and if a 0 bit and a 1 bit are equally likely is;
so ; if bits string should not contain a 0 and all other places should be occupied by 1; Then:
;
Answer:
B
Step-by-step explanation:
Data set D does not contain the value 128, which is the median value.
Data set C does not contain the outlier value 91.
Data set A contains value 168, which does not show up on the plot.
The only remaining choice is B.
_____
In order, the data values of set B are ...
... 91, 114, 120, 126, 128, 128 134, 136, 139, 142, 152
The median value of these 11 is the 6th one: 128. The median values of the remaining two sets of 5 are 120 and 139, making these values the quartiles at the ends of the box. The value 91 is more than 1.5 times the IQR (19) below the 1st quartile, so is considered an outlier. (The cutoff is 120-1.5·19=91.5.)
Y=24x+48
24 per hour (x) plus the 48 she already made would equal a total (y).
Answer:
All Real Numbers
Step-by-step explanation:
The domain is all the x values for a function. As this function is not restricted in any way, the domain is all real numbers. You can put any number in for x and plot it on the graph.
It is a prime because the only other factor of 6w^3-13 is 1.
