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Sholpan [36]
3 years ago
8

Can you help me with this?

Mathematics
1 answer:
Firdavs [7]3 years ago
3 0
I don’t know this answer but I think it’s 81
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What is this percent in simplest form? 15 3/5%=
Paul [167]

Answer:

0.156

Step-by-step explanation:

15 3/5%=(15+ 0.6)/100=15.6/100=0.156

5 0
3 years ago
A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed
tresset_1 [31]

Answer:

a) z= \frac{34-34}{2.5}= 0

z= \frac{39-34}{2.5}= 2

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) P(X

z= \frac{31.5-34}{2.5}= -1

So one deviation below the mean we have: (100-68)/2 = 16%

c) z= \frac{29-34}{2.5}= -2

z= \frac{36.5-34}{2.5}= 1

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

X \sim N(\mu = 34, \sigma=2.5)

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

P(34 < X

We can find the number of deviation from the mean with the z score formula:

z= \frac{X -\mu}{\sigma}

And replacing we got

z= \frac{34-34}{2.5}= 0

z= \frac{39-34}{2.5}= 2

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

P(X

z= \frac{31.5-34}{2.5}= -1

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

P(29 < X

And replacing we got:

z= \frac{29-34}{2.5}= -2

z= \frac{36.5-34}{2.5}= 1

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

7 0
3 years ago
The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probabili
Mariana [72]

Answer:

a) 0.01111

b) 0.4679

c) 0.33747

Step-by-step explanation:

We are given the following in the question:

The number of accidents per week can be treated as a Poisson distribution.

Mean number of accidents per week = 4.5

\lambda= 4.5

Formula:

P(X =k) = \displaystyle\frac{\lambda^k e^{-\lambda}}{k!}\\\\ \lambda \text{ is the mean of the distribution}

a) No accidents occur in one week.

P(x =0)\\\\= \displaystyle\frac{4.5^0 e^{-4.5}}{0!}= 0.01111

b) 5 or more accidents occur in a week.

P( x \geq 5) = 1-\displaystyle \sum P(x

c) One accident occurs today.

The mean number of accidents per day is given by

\lambda = \dfrac{4.5}{7} = 0.64

P(x =1)\\\\= \displaystyle\frac{0.64^1 e^{-0.64}}{1!}= 0.33747

5 0
3 years ago
Write the prime factorization of each number with factoring in increasing order of the base, using exponents for
ale4655 [162]
I don’t know sorry!!!
8 0
3 years ago
Jane will receive 18% less of her regular pay when she retires. her regular pay is $500 per week.
kirza4 [7]

0.82 \times 500 =410
7 0
3 years ago
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