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Stella [2.4K]
3 years ago
8

Males have red blood cell counts with a mean of 4.719 and a standard deviation of .490, while females have red blood cell counts

with a mean of 4.349 and a standard deviation of .402. Who has the higher count relative to the sample from which it came, the male with a count of 5.58 or a female with a count of 5.23
Mathematics
1 answer:
iVinArrow [24]3 years ago
7 0

Answer:

  • <em><u>The female  has a higher count relative to the sample of females than the male relative to the sample of males.</u></em>

Explanation:

When you want to compare the means of two samples with different statistics you may use the z-score.

The z-score measures how far a value (data point) is from the mean, in terms of the standard deviation.

The z-score is calculated as:

          z-score=(x-\mu )}/\sigma

Where:

          x=\text{value of the data}\\\\\mu =mean\\\\\sigma = \text{standard deviation}

<u>1. Male with a count of 5.58 </u>

       z-score=(5.58-4.719)/0.490=1.757

<u />

<u>2. Female with a count of 5.23</u>

     z-score=(5.23-4.349)/0.402=2.192

Therefor, the z-score of the female is greater than the z-score of the male.

The greater z-score means that, relative to her group, the female has a higher count than the male.

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D

Step-by-step explanation:

complementary sums until 90 degrees

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Solve 3^2x =30<br><br> a- log3 (30)/2<br> b- log3 15<br> c- 2(log 3 30)<br> d- log 3 60
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2 years ago
A cell phone provider classifies its customers as Low users (less than 400 minutesper month) or High users (400 or more minutes
Svet_ta [14]

Answer:

a ) See step by step explanation

b) 50%   and  50%

Step-by-step explanation:

The definition of type of user x is:

if x < 400 minutes  per month  low user

if x > 400 minutes per month  high user

The value for these states are of course yes  or no  ( the user is low or high user)

Then:

if the user is low today  xt  (0)    we have the probability of 80% ( 0,8) of have the user in the same condition and 20% (0,2) of having him or her as high user

xt = 1      

If the user is high today xt (1) is high today we have the probability of 70% ( 0,7) of having  the user in the same condition and 30% (0,3) of having him or her as high user

a) Then we construct the stochastic matrix

 Type of user

         (xt)               xt         xt+1  

      Low (0) or L   0,8        0,2

     High(1)  or H   0,3        0,7

The resulting equation system is:

xt  =  0,8* xt  +  0,2* xt+1

xt+1  = 0,3*xt + 0,7* xt+1

1  =  xt + xt+1

Solving for variables         xt+1  =  1  - xt

xt  = 0,8*xt + ( 1 - xt ) * 0,2

xt - 0,8*xt = 0,2 - 0,2xt

xt - 0,6*xt = 0,2

0,4*xt = 0,2

xt = 0,5       and     xt+1  = 0,5

b)  If in January 0,5  ( 50%) of customesr were  low user

In february  teh % of low user customer wll be

xt  =  0,8* xt  +  0,2* xt+1

0,5 = 0,8*0,5 + 0,2*xt+1

0,5 = 0,4 + 0,2 *xt+1

0,1/0,2 = xt+1        xt+1 = 0,5       or   50%

The same answer for March

7 0
3 years ago
Consider the differential equationy′′+3y′−10y=0.(a) Find the general solution to this differential equation.(b) Now solve the in
RSB [31]

Answer:

Y= 2e^(5t)

Step-by-step explanation:

Taking Laplace of the given differential equation:

s^2+3s-10=0

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Hence, the general solution will be:

Y=Ae^(-2t)+ Be^(5t)………………………………(D)

Put t = 0 in equation (D)

Y (0) =A+B

2 =A+B……………………………………… (i)

Now take derivative of (D) with respect to "t", we get:

Y=-2Ae^(-2t)+5Be^(5t)   ....................... (E)

Put t = 0 in equation (E) we get:

Y’ (0) = -2A+5B

10  = -2A+5B ……………………………………(ii)

2(i) + (ii) =>

2A+2B=4 .....................(iii)

-2A+5B=10 .................(iv)

Solving (iii) and (iv)

7B=14

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Now put B=2 in (i)

A=2-2

A=0

By putting the values of A and B in equation (D)

Y= 2e^(5t)

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3 years ago
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