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3 years ago

210,704 in equivalent Form

Mathematics

1 answer:

Verizon [17]3 years ago

4 0

Answer:

<h2><em>210= 2×3×5×7 (4 distinct prime factors)</em></h2><h2><em>2×2x2x2x2x2x11 or 2^6×11 (7 prime factors, 2 distinct)</em></h2><h2><em>please mark me as braini</em><em>est </em></h2>

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X^2+2y=19<br> X^2+y^2=34

Katen [24]

Answer:

x=3

y=5

Step-by-step explanation:

3×3=9

5×5=25

9+25=34

[3^2+5^2=34]

3^2=9

2×5=10

9+10=19

[3^2+2×5=19]

5 0

3 years ago

the percent of sales tax is 10%. what is the actual price of a pair of basketball shoes that cost $98?

larisa [96]

Answer:

(98/100)*110 = $107.8

Step-by-step explanation:

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3 years ago

4 = -8Z - 7 + 8z<br> Giving the brainliest to whoever gets the right answer.

liubo4ka [24]

Answer:

No Solution

Step-by-step explanation:

if one side is 4, another side -8z and +8z cancels, left -7

4 ≠ -7

8 0

3 years ago

Read 2 more answers

Distributive property 27+36

UkoKoshka [18]

Answer:

I think it will be equal to 63

Step-by-step explanation:

27+46=63

3 0

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

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

Finally, remember the rule to negate events:

$P(E) = 1-P(\lnot E)$

So, if we lose with probability 0.39, we win with probability

$1-0.39 = 0.61$

6 0

3 years ago