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2 years ago

To rent a machine for 6 hours, it costs $27. How much will it cost to rent it for 3 hours?

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

8 0

Answer:

$18.5

Step-by-step explanation:

6 divided by 3 = 2

27 divided by 2 = 18.5

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Solving Quadratic Equations A B C D

mixas84 [53]

I believe that it’s B.

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3 years ago

Amadahy wants to buy a card and two gifts for her friend. Cards cost 3.49 each and gifts cost $10.63 each. Amadahy has a 20$ bil

BartSMP [9]

Answer:

Given that Amadahy has a $ 20 bill, and wants to buy a card and two gifts for her friend, and knowing that each card costs $ 3.49 and each gift costs $ 10.63, the total cost of the 3 items would be $ 24.75 (10.63 x 2 + 3.49 ).

Therefore, she will not be able to purchase the products she wants, since she will need $ 4.75 to reach the amount necessary for it (24.75 - 20).

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3 years ago

Determine all the possible orders if someone went into a bakery and could order a vanilla, strawberry, or chocolate cake. They c

AysviL [449]

Answer:

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answer is C = 12

5 0

3 years ago

What is the value of StartFraction 1.6 times 10 Superscript 14 Baseline Over 3.2 times 10 Superscript 7 Baseline EndFraction in

Helga [31]

Answer:

5.0 times 10 Superscript 6

Step-by-step explanation:

What is the value of StartFraction 1.6 times 10 Superscript 14 Baseline Over 3.2 times 10 Superscript 7 Baseline?

This is represented mathematically as:

1.6 × 10¹⁴/ 3.2 × 10⁷

Solving for this

1.6 × 10¹⁴/ 3.2 × 10⁷

= [1.6/3.2] × 10¹⁴-⁷

= 5 × 10⁶

= 5.0 times 10 Superscript 6

4 0

2 years ago

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

3 years ago