Answer:
m = 1
Step-by-step explanation:
We can suppose that the number we are looking for is for example 5.
(we can do so because the probability is the same for each number - it'sna fair dice)
For the first toss the probability we have 5 is 1/6 (we have 6 numbers on the dice and number 5 is just one of the possible 6 outcomes).
For the second toss the probability we have 5 is again 1/6.
For the rest of 3 tosses we don'tcare what number we will get( we have our two consecutive 5s), so all of the outcomes for the rest of 3 tosses are good for us (probability is 6/6 = 1)
Threfore, the probability to get two consecutive 5s is 1/6 * 1/6 * 1 * 1 * 1 = 1/36.
We can see that m = 1.
For number 1, the median is 8
For number 2, the mode is C.100
And for number 3, the mean is D.410
Answer: Six places to the right
Step-by-step explanation:
For this exercise it is important to remember that. by definition, the exponent of a number indicates the number of times you must use the same factor to multiply.
Given:

"b" is the base and "n" is the exponent.
In this case, you have the following multiplication provided in the exercise:

Notice that the base 10 has an exponent 6. This indicates the following:

By definition, moving the decimal point 6 places to the right (because there are six zeros), is the same as multiplying the decimal number 34.05 by 1,000,000.
Therefore, based on the explained, when you move the decimal point six places to the right, you get the following product.

You can use at most 24 beads that are 10 inches long to decorate the rope.
Step-by-step explanation:
Given,
Length of rope = 120 feet
1 foot = 12 inches
20 foot = 20*12 = 240 inches
Length of bead = 10 inches
Let,
x represents the number of beads used on rope.
As the rope is 240 inches long, therefore, the total length of beads cannot exceed 240 inches.
Length of each bead*Number of beads ≤ length of rope

Dividing both sides by 10

You can use at most 24 beads that are 10 inches long to decorate the rope.
Keywords: inequality, division
Learn more about inequalities at:
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