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Morgarella [4.7K]
3 years ago
12

2.5(6v-4)=10+4(1.5+0.5)

Mathematics
1 answer:
asambeis [7]3 years ago
8 0
2.5(6v-4)=10+4(1.5+0.5)\\\\(2.5)(6v) - 2.5(4)=10 + 4(1.5) + 4(0.5)\\\\15v - 10 = 10 + 6+2 \\\\15v - 10 = 18\\\\15v - 10 + 10 = 18 + 10\\\\15v = 18\\\\\frac{15v}{15} =\frac{18}{15} \\\\v = \frac{18}{15}\\\\v = \frac{6\times 3}{5\times 3}\\\\v=\frac{6}{5}\\\\\boxed{\bf{ v = \frac{6}{5} = 1\frac{1}{5}}}
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Answer:

C.

Step-by-step explanation:

(6^7)^3 means (6^7)(6^7)(6^7).

When multiply numbers with same base, add the exponents.

6^(7+7+7)=6^(3*7)=6^(21).

In the beginning you could have just multiply 7 and 3 so the answer is 6^(21).

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L. Find the positive square root of each of the following numbers:(a) 36(b) 81(d) 256(c) 196
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2 years ago
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gtnhenbr [62]

Answer:

a) 0.5476

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c) 0.4524

Step-by-step explanation:

Given this information, we can conclude that 74% of the cars won't need any repairs over a 1-year period (100 - 20 - 5 - 1 = 74%). And 26% will need at least 1 repair over a 1-year period.

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We need that car 1 won't need repair AND car 2 won't need repair.

=P(Car 1 doesn't need repair) x P(Car 2 doesn't need repair)

= 0.74 x 0.74 = 0.5476

The probability that neither will need repair is 0.5476.

<u>b) Both will need repair:</u>

We need that car 1 needs repair AND car 2 needs repair.

P(Car 1 needs repair) x P(Car 2 needs repair)

= 0.26 x 0.26 = 0.0676

The probability that both will need repair is 0.0676

<u>c) At least one car will need repair</u>

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To solve this one, it's easier to use the complement of P(neither needs repair)

1 - P(neither needs repair)

1 - (0.74)(0.74)  = 1 - 0.5476 = 0.4524

The probability that at least one car will need repair is 0.4524

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3 years ago
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vovikov84 [41]

Answer:

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