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andrew11 [14]
4 years ago
9

Convert 12 inches to millimeters (to the nearest tenth).

Mathematics
2 answers:
Novosadov [1.4K]4 years ago
8 0
12 inches to millimeters is 305.0 if u put it the nearest tenth
Mnenie [13.5K]4 years ago
4 0

304.8

is the one with out rounding it up


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HURRY IM TIMED!!!
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Answer:

A, C, and D

Step-by-step explanation:

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Reuben would like to buy sports car that cost $35,000 today when he graduates from college in five years if the rate of informat
alexandr402 [8]

Answer:

<h2>$40,250</h2>

Step-by-step explanation:

Amount he will pay in the next five years = Principal + Interest

Given Principal =  $35,000

Simple Interest = PRT/100

R is the rate = 3%

T is the time in years = 5years

SI = 35,000 * 3 * 5/100

SI =  $5,250

Amount =  $35,000  +  $5,250

Amount =  $40,250

Ruben will have to pay $40,250 after 5years

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4 years ago
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Rectangle abcd is similar to rectangle pqrs given that ab =14cm bc =8cm and pq=12cm calculate the length of qr
ohaa [14]

Answer: 6.8571 (Round as needed)

Hope this is correct

Step-by-step explanation:

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Using the similarity statement we can say 14:12, (ab:pq)

That is our ratio.

So we do 14/12 = 8/x, we solve this using algebra to get our answer.

Hope this is correct.

7 0
3 years ago
The probability is 1 in 4,011,000 that a single auto trip in the United States will result in a fatality. Over a lifetime, an av
IrinaVladis [17]

Answer:

0.0114

Step-by-step explanation:

(a) What is the probability of a fatal accident over a lifetime?

Suppose A be the event of a fatal accident occurring in a single trip.

Given that:

P(1 single auto trip in the United States result in a fatality) = P(A)

Then;

P(A) = 1/4011000

P(A) = 2.493 × 10⁻⁷

Now;

P(1 single auto trip in the United States NOT resulting in a fatality) is:

P(\mathbf{\overline A}) = 1 -  P(A)

P(\mathbf{\overline A}) = 1 - 2.493 × 10⁻⁷

P(\mathbf{\overline A}) = 0.9999997507

However, P(fatal accident over a lifetime) = P(at least 1 fatal accident in lifetime i.e. 46000 trips)

= 1 - P(NO fatal accidents in 46000 trips)

Similarly,

P(No fatal accidents over a lifetime) = P(No fatal accident in the 46000 trips) = P(No fatality on the 1st trip and No fatality on the 2nd trip ... and no fatality on the 45999 trip and no fatality on the 46000 trip)

= [P(\overline A)] ^{46000}  \ \ \  (since \  trips \ are \ independent \ events)

= [0.9999997507]^{46000}

= 0.9885977032

Finally;

P(fatal accident over a lifetime) = 1 -  0.9885977032

P(fatal accident over a lifetime) = 0.0114022968

P(fatal accident over a lifetime) ≅ 0.0114

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Answer:

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