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gregori [183]
3 years ago
12

Colin wants to set up an aquarium.. He already has a tank, but needs to purchase fish, filters, and plants. If the cost of the f

ish
is $84.79, the filters usually cost $60.45 but Colin finds them on sale for $44.75, and the cost of the plants is $18.66, determine
how much of Colin's $488.85 savings will be used to set up his aquarium.
a $289.81
Mathematics
2 answers:
ahrayia [7]3 years ago
7 0

Answer:

340.65

Step-by-step explanation: Add 84.79+44.75+18.66=148.2

148.2-$488.85= 340.65

Kisachek [45]3 years ago
7 0

Answer:

$148.2

Step-by-step explanation:

He needs to purchase fish, filters, and plants

Cost of the fish: $84.79

Cost of the filters: $44.75

Cost of the plants: $18.66

Total cost: $84.79 + $44.75 + $18.66 = $148.2

Despite how much Colin saved, he spent $148.2 to set up an aquarium.

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There are eight different jobs in a printer queue. Each job has a distinct tag which is a string of three upper case letters. Th
Vikentia [17]

Answer:

a. 40320 ways

b. 10080 ways

c. 25200 ways

d. 10080 ways

e. 10080 ways

Step-by-step explanation:

There are 8 different jobs in a printer queue.

a. They can be arranged in the queue in 8! ways.

No. of ways to arrange the 8 jobs = 8!

                                                        = 8*7*6*5*4*3*2*1

No. of ways to arrange the 8 jobs = 40320 ways

b. USU comes immediately before CDP. This means that these two jobs must be one after the other. They can be arranged in 2! ways. Consider both of them as one unit. The remaining 6 together with both these jobs can be arranged in 7! ways. So,

No. of ways to arrange the 8 jobs if USU comes immediately before CDP

= 2! * 7!

= 2*1 * 7*6*5*4*3*2*1

= 10080 ways

c. First consider a gap of 1 space between the two jobs USU and CDP. One case can be that USU comes at the first place and CDP at the third place. The remaining 6 jobs can be arranged in 6! ways. Another case can be when USU comes at the second place and CDP at the fourth. This will go on until CDP is at the last place. So, we will have 5 such cases.

The no. of ways USU and CDP can be arranged with a gap of one space is:

6! * 6 = 4320

Then, with a gap of two spaces, USU can come at the first place and CDP at the fourth.  This will go on until CDP is at the last place and USU at the sixth. So there will be 5 cases. No. of ways the rest of the jobs can be arranged is 6! and the total no. of ways in which USU and CDP can be arranged with a space of two is: 5 * 6! = 3600

Then, with a gap of three spaces, USU will come at the first place and CDP at the fifth. We will have four such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 4 * 6!

Then, with a gap of four spaces, USU will come at the first place and CDP at the sixth. We will have three such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 3 * 6!

Then, with a gap of five spaces, USU will come at the first place and CDP at the seventh. We will have two such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 2 * 6!

Finally, with a gap of 6 spaces, USU at first place and CDP at the last, we can arrange the rest of the jobs in 6! ways.

So, total no. of different ways to arrange the jobs such that USU comes before CDP = 10080 + 6*6! + 5*6! + 4*6! + 3*6! + 2*6! + 1*6!

                    = 10080 + 4320 + 3600 + 2880 + 2160 + 1440 + 720

                    = 25200 ways

d. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways. Similarly, if LPW comes last, the remaining 7 jobs can be arranged in 7! ways. so, total no. of different ways in which the eight jobs can be arranged is 7! + 7! = 10080 ways

e. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways in the queue. Similarly, if QKJ comes second-to-last then also the jobs can be arranged in the queue in 7! ways. So, total no. of ways to arrange the jobs in the queue is 7! + 7! = 10080 ways

5 0
4 years ago
The length of Dwight's garden is 12.5 meters. Dwight measured it but made a measurement error of 6%. Find the lowest and highest
pishuonlain [190]

Answer: See Explanation

Step-by-step explanation:

From the question, The length of Dwight's garden is 12.5 meters but he measured it and made a measurement error of 6%.

The lowest measurement Dwight could have made will be:

= 12.5 - (6% × 12.5)

= 12.5 - (0.06 × 12.5)

= 12.5 - 0.75

= 11.75 meters

The lowest measurement Dwight could have made will be:

= 12.5 + (6% × 12.5)

= 12.5 + (0.06 × 12.5)

= 12.5 + 0.75

= 13.25 meters

3 0
3 years ago
Help me please !!!!!
andrew-mc [135]

Answer:

Step-by-step explanation:

The way to do this is just to add the two real parts together and add the two imaginary parts together.

2 + 5j + 7 - 3j

2 + 7 + 5j - 3j

9 + 2j

3 0
3 years ago
In Justin's school 0.825 of the students participate in a sport. If there are 1000 students in Justin's school, how many partici
ikadub [295]

Answer:

825

Step-by-step explanation:

Let the number of students who participated in the sport be x, which is equivalent to 0.825

The total number of students in the school is equivalent to 1

Expressing mathematically,we obtain

0.825:x=1:1000

\implies  \frac{0.825}{x}= \frac{1}{1000}

By cross multiplying we obtain,

x=0.825 \times 1000

We finally simplify to arrive at

x=825

Hence the number of students who participated in the sport is 825

8 0
4 years ago
Read 2 more answers
Pwease answer dis for me pweez, show work too..... :((
Lostsunrise [7]

Answer:

It does show direct variation.

Step-by-step explanation:

Every hour, you receive $8.50. This doesn't change throughout the entire table. On a graph, this would look like a line going through the origin, with points at (1, 8.5). (2, 17), (3, 25.5), and so on.

5 0
3 years ago
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