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rodikova [14]
3 years ago
5

A bike rental company charges $60 for the first hour, and $25 for each additional hour. Michael wants to spend less than $200 on

his bike rental for the day. What is the maximum number of whole hours that Michael can rent a bike? Which inequality and answer correctly represents the above situation? A) 60 + 25h > 200; 5 whole hours B) 25 + 60h < 200; 6 whole hours C) 25 + 60h > 200; 5 whole hours Eliminate D) 60 + 25h < 200; 5 whole hours
Mathematics
2 answers:
vladimir1956 [14]3 years ago
6 0
It is d. The h is the amount of hours and this must be paired with 25 because that is the rate do you can eliminate b and c. It is not a because that is making it so the cost is above 200 when you are trying to keep it below that
-Dominant- [34]3 years ago
4 0
It is d. The h is the amount of hours and this must be paired with 25 because that is the rate do you can eliminate b and c. It is not a because that is making it so the cost is above 200 when you are trying to keep it below that
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If the customer rolls the dice and rents a second movie every Thursday for 20 consecutive weeks, what is the total amount that t
VMariaS [17]

Answer:

The expected amount for a duration of 20 weeks is $9.4

<em></em>

Step-by-step explanation:

Read attachment to understand the complete question

From the question, we have the following given parameters:

x = Amount paid for a second movie

E(x) = 0.47 --- Expected value of x

SD(x) = 0.15 --- Standard deviation of x

Required

Determine the total amount paid for a second movie for 20 weeks

The required implies that we calculate the expected value for a duration of 20 weeks.

i.e.

n = 20

If

E(x) = 0.47

Then:

E(nx) = n * E(x)

Substitute 20 for n

E(20x) = 20 * E(x)

Substitute 0.47 for E(x)

E(20x) = 20 * 0.47

E(20x) = 9.4

<em>Hence, the expected amount for a duration of 20 weeks is $9.4</em>

6 0
3 years ago
In this exercise, we consider strings made from uppercase letters in the English alphabet and decimal digits.
Yuri [45]

Answer:

  • a) 26^2 36^8
  • b) 21\cdot10\cdot36^7
  • c) 5^3 31^7
  • d) 10\cdot 9\cdot 8 \cdot 7 \cdot 26^6

Step-by-step explanation:

We will use the product rule from combinatorics.

  • a) There are 26 letters in the English alphabet, so there are 26 possible choices for the first character and 26 possible choices for the last one. Each one of the remaining eight characters of the string has 36 choices (letters or digits). By the product rule, there are 26\cdot36\cdot 36\cdots 36\cdot 26=26^2 36^8 strings.
  • b) We have 5 possible choices for the first character, it must be some vowel a,e,i,o,u. The second character can be chosen in 21 ways, selecting some consonant. There are 10 possibilities for the last character because only of the digits are allowed. The other seven characters have no restrictions, so each one can be chosen in 36 ways. By the product rule there are 21\cdot 10\cdot 36^7 strings.    
  • c) The third character has 5 possibilities. Repetition of vowels is allowed, so the sixth and eighth characters have each one 5 possible choices. There are seven characters left. None of them are a vowel, but they are allowed to take any other letter or digit, so each one of them can be chosen in 36-5=31 ways. Therefore there are 5^3 31^7 strings.
  • d) Remember that the binomial coefficient \binom{n}{k} is the number of ways of choosing k elements from a set of n elements. In this case, to count all the possible strings, we first need to count in how many ways we can select the four positions that will have the digits. This can be done in \binom{10}{4} ways, since we are choosing four elements from the set of the ten positions of the string. Now, for the first position, we can choose any digit so it has 10 possibilities. The second position has 9 possibilities, because we can't repeat the digit used on the first position. Similarly, there are 8 choices for the third position and there are 7 choices for the fourth. Now, these are the only digits on the string, so the remaining 6 characters must be letters, then each one of them has 26 possibilities. By the product rule, there are 10\cdot 9\cdot 8 \cdot 7 \cdot 26^6 strings.
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