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3 years ago

Which fraction is closer to 0 than 1

Mathematics

1 answer:

scoundrel [369]3 years ago

3 0

Answer:

5/12

Step-by-step explanation:

Since 1/2 is the same distance from zero and one, we can use it to judge whether a fraction is closer to 0 or 1.

Since 5/8 is greater than 4/8 which is 1/2, it is closer to 1.

Since 8/10 is greater than 5/10 which is 1/2, it is closer to 1.

Since 5/12 is less than 6/12 which is 1/2, it is closer to 0.

So our answer is C. 5/12

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2 years ago

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ryzh [129]

Answer:

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Step-by-step explanation:

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2 years ago

A toll-free sales line sold 85 products for every 125 calls in one day. What is the daily success rate of the sales line? A. 0.1

IgorLugansk [536]

Answer:

C. 0.68

Step-by-step explanation:

Given;

number of products sold in a day by toll-free sales line = 85 products

number of calls in a day = 125

The daily success rate of the sales line is given by the ratio of the total products in a day to total number of calls in a day.

The daily success rate of the sales line = total products sold / number of calls

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7 0

2 years ago

Recall that for use of a normal distribution as an approximation to the binomial distribution, the conditions np greater than or

vagabundo [1.1K]

Answer:

The minimum sample size needed for use of the normal approximation is 50.

Step-by-step explanation:

Suitability of the normal distribution:

In a binomial distribution with parameters n and p, the normal approximation is suitable is:

np >= 5

n(1-p) >= 5

In this question, we have that:

p = 0.9

Since p > 0.5, it means that np > n(1-p). So we have that:

$n(1-p) \geq 5$

$n(1 - 0.9) \geq 5$

$0.1n \geq 5$

$n \geq \frac{5}{0.1}$

$n \geq 50$

The minimum sample size needed for use of the normal approximation is 50.

6 0

2 years ago

Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

7 0

2 years ago

Which fraction is closer to 0 than 1​

Which fraction is closer to 0 than 1