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kodGreya [7K]
2 years ago
12

1 1/2 is_____% of 7 1/2

Mathematics
1 answer:
Xelga [282]2 years ago
5 0

Answer:

1 1/2 is__11/15___% of 7 1/2

Step-by-step explanation:

Given

Number - 7\frac{1}{2}  = \frac{15}{2}

Let X % of \frac{15}{2} = \frac{11}{2}

X % is thus equal to

\frac{11}{2} * \frac{2}{15} \\\frac{11}{15}

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The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day
Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of  length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(between 236 and 281 days)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%

b) a) P(last between 236 and 296)

P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least 1-\dfrac{1}{k^2}  data lies within k standard deviation of mean.

For k = 2

1-\dfrac{1}{(2)^2} = 75\%

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

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yanalaym [24]

Answer:

There is not enough information to answer this question

Step-by-step explanation:

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Step-by-step explanation:

5x−4=5

1 Add 44 to both sides.

5x=5+4

5x=5+4

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3 Divide both sides by 55.

x=\frac{9}{5}

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Answer:

y = 3x

Step-by-step explanation:

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