Which statements about the composite figure are true? Check all that apply.

Mathematics

1 answer:

AnnyKZ [126]3 years ago

5 0

Where’s the answers ?

Step-by-step explanation:

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Ms. Coleman's class has already collected 403 cans for a food drive, but the class wants to collect a total of at least 1,000 ca

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We need to figure out how many more cans they need to bring. We figure this out by subtracting 403 from 1,000, which gives us 596 more cans. We now divide 596/28, to find how many each student would need to bring. This equals 21.3. You might want to mention that each student can't bring .3 of a can, so the answer could be 22 also.

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3 years ago

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Train a leaves new york at 9am eastern time on monday, headed for los angeles at a constant rate of 40 miles per hour. on the sa

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We can assume that B leaves Los Angeles not at noon but at 2pm Eastern Time, so 5 hours later than A leaves New York.

In 5 hours A covers 5*40=200 miles. So, A and B together should cover 3,000-200=2,800 miles.

Combined rate = 40 + 60 = 100 miles per hour.

To cover 2,800 miles they'll need 2,800/100=28 hours.

2pm Eastern Time on Monday + 28 hours = 6pm Eastern Time on Tuesday.

Answer: 6:00pm on Tuesday

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3 years ago

LINEAR ALGEBRA<br><br> List four variables and link them using one equation

Studentka2010 [4]

A+b+c+d=18 is an example

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3 years ago

A small airline overbooks flights on the assumption that several passengers will not show up. Suppose that the probability that

olchik [2.2K]

We have 22 tickets sold, and 20 seats. This means that at least 2 passengers must not show up (otherwise, at least 21 passengers will be present, and there wouldn't be space for them).

Considering each passenger as independent, you can think of this experiment. Suppose you toss a coin for each passenger. If the coin lands on heads, the passenger shows up. If it lands on tails, the passenger doesn't show up.

But the coin is unfair: it has a 0.91 probability of landing on heads, and thus 0.09 probability of landing on tails.

This implies that the probability of having exactly $k$ tails is

$\binom{22}{k} \cdot 0.09^k \cdot 0.91^{22-k}$

We already concluded that at least two passengers must not show up. So, if our coins lands on tails less than twice, we've lost. So, the losing probability is

$\displaystyle\binom{22}{0}\cdot 0.09^0 \cdot 0.91^{22} + \binom{22}{1}\cdot 0.09^1 \cdot 0.91^{21} \approx 0.39$