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Nezavi [6.7K]
3 years ago
12

5.3: Solutions on a Number Line

Mathematics
1 answer:
andrew11 [14]3 years ago
5 0

Answer:

idfk

Step-by-step explanation:

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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.

7 0
3 years ago
How do you solve it?
Elis [28]

At heart we're being asked for a line through two points,


(40^\circ \textrm{ C}, 355 \textrm { m/s}) \quad \textrm{and} \quad (49^\circ \textrm{ C}, 360 \textrm { m/s})


In general the line through (a,b) and (c,d) is


(y-b)(c-a)=(x-a)(d-b)


Check that you understand why both (a,b) and (c,d) are on this line.


Here our indepedent variable, instead of x, is T, temperature. Our dependent variable is v, velocity. Substituting,


(v - 355)(49 - 40) = (T - 40)(360 - 355)


9(v - 355) = 5(T - 40)


v-355 = \frac 5 9 T - \frac{200}{9}


v= \frac 5 9 T - \frac{200}{9} + 355


v= \frac 5 9 T + \frac{2995}{9}


That's our answer; let's check it.


When T=40, v = (5/9)40 + (2995/9) = 355 good


When T=49, v= (5/9)49 + (2995/9) = 360 good




7 0
3 years ago
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