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Anna007 [38]
3 years ago
5

13. Add [-6-2 2] + [-3 2 1]. A. (-9-3-5] B. [-3 0 3] C. [-6 0 3] D. [-9 0 3]

Mathematics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer: The answer should be C if you were to add each point.

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396 divided by 88 with work shown
RSB [31]
The answer is 4.5, I hope this helps.

6 0
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Express the repeating decimal 2.7 as a fraction.
sergeinik [125]
2/9 is the answer broski
8 0
2 years ago
a truck is rented at 40$ per day plus a charge per mile use. The truck traveled 15 miles in one day, and the total charge was 11
Alla [95]

Answer:

15x + 40 = 115

We set the equation to 115 because the total cost is $115.

We add the "+ 40" because of the $40 to rent the trailer for the day.

The 15x is because it represents the miles per hour (x) times the number of hours.

The number of hours times the charge per mile, plus the $40 to rent it, will equal $115.

Hope this helps!!

5 0
3 years ago
Statistics question about random probability Cheese pastureized or Raw MilkA cheese can be classified as either raw-milk or past
blsea [12.9K]

Answer:

(a) Two cheeses are chosen at random. the probability that both cheeses are pasteurized is Pr(PP) = 0.82 x 0.82 = 0.6724 to 4 decimal places)

(b) Four cheeses are chosen at random. The probability that all four cheeses are pasteurized is Pr(PPPP) = 0.82 x 0.82 x 0.82 x 0.82 = 0.4521 to 4 decimal places

(c) What is the probability that at least one of four randomly selected cheeses is raw-milk is Pr(RPPP) Or Pr(RRPP) Or Pr(RRRP) Or Pr(RRRR)

= 0.1269 to 4 decimal places

It would not be unusual that at least one of four randomly selected cheeses is raw-milk, because the probability have a value between 0 and 1

Step-by-step explanation:

If is given that 80% of the cheese is classified as pasteurized.

It then implies that 20% of the cheese is classified as Raw-milk

Probability of pasteurized cheese is 0.82(Denoted by Pr(P))

Probability of raw-milk cheese is 0.18(Denoted as Pr(R))

(a) Two cheeses are chosen at random. the probability that both cheeses are pasteurized is Pr(PP) = 0.82 x 0.82 = 0.6724 to 4 decimal places)

(b) Four cheeses are chosen at random. The probability that all four cheeses are pasteurized is Pr(PPPP) = 0.82 x 0.82 x 0.82 x 0.82 = 0.4521 to 4 decimal places

(c) What is the probability that at least one of four randomly selected cheeses is raw-milk is Pr(RPPP) + Pr(RRPP) + Pr(RRRP) + Pr(RRRR)

(0.18 x 0.82 x 0.82 x 0.82) + (0.18 x 0.18 x 0.82 x 0.82) + (0.18 x 0.18 x 0.18 x 0.82) + (0.18 x 0.18 x 0.18 x 0.18) = 0.1269 to 4 decimal places

3 0
3 years ago
Here is Laila’s graph from the previous screen.
lidiya [134]

Answer:

y=-1/3x+8

Step-by-step explanation:

There is no need for any specific answers, but here is one that could logically work out. Since the graph is going left/down, it has a negative slope, so -1/3 would be reasonable. The graph doesn't cross the origin and crosses above it, so this equation must have a positive 'b' value. In this case, I chose 8. y=-1/3+8 could represent Laila's graph.

4 0
2 years ago
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