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solniwko [45]
3 years ago
10

Write the prime factorization of 36. Use exponents when appropriate and order the factors from least to greatest (for example, 2

235).
Mathematics
1 answer:
sladkih [1.3K]3 years ago
6 0

Answer:

                                 36

                        2        x       18

                  2      x       2       x   9

             2   x     2    x    2     x      3

Step-by-step explanation:

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Choose the correct simplification of (7x3y3)2.<br><br> 14x6y6<br> 49x5y5<br> 49x6y6<br> 14x5y5
dolphi86 [110]
The answer would be the first one.

I find it easier to expand the equation (you still get the same answer).

= 2 * 7x + 2 * 3y + 2 * 3
= 14x + 6y + 6

Hope this helps!
8 0
3 years ago
Read 2 more answers
Y=4x+1 complete the table
Tom [10]

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3 0
2 years ago
During a flu epidemic, 35% of the school's students have the flu. Of those with the flu, 90% have high
frutty [35]

Answer: 0.8015

Step-by-step explanation:

Let F= event that a person has flu

H= event that person has a high temperature.

As per given,

P(F) =0.35

Then P(F')= 1- 0.35= 0.65               [Total probability= 1]

P(H | F) = 0.90

P(H|F') = 0.12

By Bayes theorem, we have

P(F|H)=\dfrac{P(F)\timesP(H|F)}{P(F)\timesP(H|F)+P(F')\timesP(H|F')}\\\\=\dfrac{0.35\times0.90}{0.35\times0.90+0.65\times0.12}\\\\=\dfrac{0.315}{0.315+0.078}\approx0.8015

Required probability = 0.8015

8 0
3 years ago
molly ran 1/4 of a mile in 1 minute if molly runs at that speed how long will it take her to run 4 miles
Vika [28.1K]

Step-by-step explanation:

Molly ran 1/4 of a mile in 1 minute, i.e.

1/4 mile : 1 minute

Multiply both sides by 4 to get time for one mile

1 mile : 4 minutes

Multiply by 4 again to get time for 4 miles

4 miles = 4*4 = 16 minutes.

So it takes Molly 16 minutes to run four miles.

7 0
3 years ago
Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c) The probability that exactly 1 of the next three vehicles passes is 0.189.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let <em>X</em> = number of vehicles that pass the inspection.

The probability of the random variable <em>X</em> is <em>P (X) = 0.70</em>.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

                                    =(0.70)^{3}\\=0.343

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

                                   =1-0.343\\=0.657

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

                         = P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

                              + P (Only 3rd vehicle passes)

                       =(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

                                                       + P (0 vehicles passes)

                                              =0.189+(0.30\times0.30\times0.30)\\=0.216

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let <em>X</em> = all 3 vehicle passes and <em>Y</em> = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0
3 years ago
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