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Softa [21]
3 years ago
6

PLEASE HELP

Mathematics
2 answers:
Evgen [1.6K]3 years ago
7 0

If i had to give it my best shot it would be 8yd.

adell [148]3 years ago
4 0
We know that
in a right triangle
Applying the Pythagoras Theorem
c²=a²+b²

in this problem
c=√87 yd
a=√23 yd
b=?
so
b²=c²-a²-----> b²=(√87)²-(√23)²----> b²=87-23----> b²=64----> b=8 yd

the answer is
8 yd
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Round 1.5 to one decimal place
xxMikexx [17]
If the last digit in 1.50 is less than 5, then remove the last digit.

If the last digit in 1.50 is 5 or more and the second to the last digit in 1.50 is less than 9, then remove the last digit and add 1 to the second to the last digit.

If the last digit in 1.50 is 5 or more and the second to the last digit in 1.50 is 9, then remove the last digit, make the second to last digit 0, and add 1 to the number to the left of the decimal place.

When rounding 1.50 to one decimal place we use One Decimal Place Rule #1. Therefore, the answer to "What is 1.50 rounded to 1 decimal place?" is:

1.5
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3 years ago
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Troyanec [42]
Multiply by b x h

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8 0
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Please help thx 10 poin
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TRUE OR FALSE MATH PLZ HELP:
Sveta_85 [38]

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3 0
3 years ago
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.

7 0
3 years ago
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