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Maurinko [17]
3 years ago
6

Show me how to find the answer to 6x1/3

Mathematics
2 answers:
mamaluj [8]3 years ago
7 0

Answer:

2

Step-by-step explanation:

Simplify. Remember to follow PEMDAS and the left-> right rule.

In this case, multiply the whole number (6/1) with the numerator (1)

6 x 1 = 6

Next, simplify by dividing 6 with 3:

6/3 = 2

2 is your answer.

~

Natasha2012 [34]3 years ago
3 0

Answer:

2

Step-by-step explanation:

6 times 1 is 6 and 6 divided by 3 is 2.

if the / is a fraction, then the answer is still the same.

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Although they don't tell you how many games he place, we can still solve. This is because we are adding an average to an average and therefore don't need to know totals.

12.375 + 1 = 13.375

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What is the measure of ZWX?
GaryK [48]

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2 years ago
Solve for x.<br><br> 6x2−2=13
makkiz [27]

Answer:

x = ±sqrt(5/2)

Step-by-step explanation:

6x^2−2=13

Add 2 to each side

6x^2−2+2=13+2

6x^2 = 15

Divide each side by 6

6/6x^2 = 15/6

x^2 = 5/2

Take the square root of each side

sqrt(x^2) = ±sqrt(5/2)

x = ±sqrt(5/2)

5 0
2 years ago
Read 2 more answers
What times 5 equals 70?
Viefleur [7K]
5n  =70 
5 x 14 = 70
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3 0
3 years ago
Read 2 more answers
Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
  • n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

8 0
3 years ago
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