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cupoosta [38]
1 year ago
13

A machine worked for 4hours and used 6kilowatts of electricity.What is the rate ofenergy consumed inkilowatts per hour?*Enter yo

ur answer as a decimal
Mathematics
1 answer:
Musya8 [376]1 year ago
4 0

4 hours ---> 6 kilowatts

1 hour -----> x kilowatts

\begin{gathered} 4\times x=1\times6 \\ 4x=6 \\ \frac{4x}{4}=\frac{6}{4} \\ x=\frac{3}{2}=1.5 \end{gathered}

answer:

1.5 kilowatts per hour

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8n= 24 Solving Multiplication equations algebraically. Check through substitutions​
stellarik [79]

Answer:

<h2>n = 3</h2>

Step-by-step explanation:

8n=24           <em>divide both sides by 8</em>

\dfrac{8n}{8}=\dfrac{24}{8}

n=3

Check:

8(3) = 24

24=24              <em>CORRECT</em>

6 0
4 years ago
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Motor vehicles sold to individuals are classified as either cars or light trucks (including SUVs) and as either domestic or impo
Rudik [331]

Answer:

a) Probability that vehicle is Light Truck is 0.69

b) Probability that vehicle is imported car is 0.08 or 8%.

Step-by-step explanation:

Part a)

Event that vehicle is a car is given by A. Since, the vehicles can only be classified as cars or light trucks, the event that a vehicle will be light truck will be compliment of A.

i.e. Event that vehicle is a Light Truck = A^{c}

It is given that in recent year 69% of vehicles sold were light trucks, 78% were domestic, and 55% were domestic light trucks.

So, from here we can say that, if a vehicle is randomly selected from all the vehicles the probability that it would be a light truck will be:

P(Vehicle will be Light Truck) = P(A^{c}) = 69% = 0.69

Part b)

Event that vehicle is imported is given by B. We need to find the probability the a randomly chosen vehicle is an imported car i.e. we have to find probability of occurrence of events A and B together, which will be denoted as: P(A and B)

Since, P(A^{c}) = 69% = 0.69

P(A) = 1 - P(A^{c}) = 1 - 0.69 = 0.31

It is also given that 78% of vehicles were domestic. This means, the percentage/probability of imported vehicles is:

P(B) = 1 - 0.78 = 0.22

55% of vehicles were domestic light trucks. This can be expressed as:

P(A^{c}\&B^{c}), the compliment of this event will give us P(A or B).

i.e.

P(A or B) = 1 - P(A^{c} \&B^{c}) = 1 - 0.55 = 0.45

According to the addition rule of probabilities:

P(A or B) = P(A) + P(B) - P(A and B)

Substituting the calculated values gives us:

0.45 = 0.31 + 0.22 - P(A and B)

P(A and B) = 0.08

This means, the probability that vehicle is imported car is 0.08 or 8%.

4 0
3 years ago
How to solve 9x 4 2/5 in words
Natalija [7]

Answer:

39.6

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9 x 4 2/5

9 x 22/5

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4 years ago
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