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Brums [2.3K]
3 years ago
5

Do any of you have a made-up pokemon? Give Me Name, Hight, Type, Category, And All of That Stuff.

Mathematics
1 answer:
Veronika [31]3 years ago
7 0

Answer:

I like pikachu. I don't know how to do te rest though

Step-by-step explanation:

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John made $122 in 8 hours.how much did he make per hour
NARA [144]

Answer:

$15.25

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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
The shelf S hangs from the ceiling by two ropes (1 and 2). A block of cheese B rests on the shelf.
Zolol [24]

Answer:

D

answer for khan

↑f N

↓F g,B

3 0
2 years ago
Read 2 more answers
Solve the equation for t A=s(t+r)x
tester [92]

Answer:

t = (A/sx) - r

Step-by-step explanation:

Solve for t like this:

A = s(t+r)x\\A = sx(t+r)\\\\\frac{A}{sx} = t+r\\\frac{A}{sx} - r = t\\t= \frac{A}{sx} - r

6 0
3 years ago
The mean and the median are the same.<br> 2,3,4,5
Ber [7]

Step-by-step explanation:

mean \: ( \bar  x) =  \frac{2 + 3 + 4 + 5}{4}  =  \frac{14}{4}  = 3.5 \\  \\ median \: =  \frac{3 + 4 }{2}  =  \frac{7}{4}  = 3.5 \\  \\

Thus, mean and median are same.

7 0
3 years ago
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