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Nikolay [14]
3 years ago
10

The mean incubation time for a type of fertilized egg kept at 100.7100.7â°f is 2323 days. suppose that the incubation times are

approximately normally distributed with a standard deviation of 22 daysdays. â(a) what is the probability that a randomly selected fertilized egg hatches in less than 2121 âdays? â(b) what is the probability that a randomly selected fertilized egg hatches between 1919 and 2323 âdays? â(c) what is the probability that a randomly selected fertilized egg takes over 2525 days toâ hatch?
Mathematics
1 answer:
Irina18 [472]3 years ago
5 0
In this item, we have the mean incubation period of 23 days and the standard deviation is 22 days. We use z-score to determine the unknown in each of the items.

(a) less than 21 days.
       z-score = (23 - 21) / 22 = 1/11
This translates to a percentage of 53.6%

(b) z-score for 19 days.
     z-score = (23 - 19) / 22 = 2/11
This translates to a percentage of 57.2%.
   
    z-score for 23 days.
   z-score = (23 - 23)/ 22 = 0
This translates to a percentage of 50%.

The difference between the two numbers is only 7.2%

(c) z-score of 25.
     z-score = (23 - 25)/ 22 = -1/11
This translates to a percentage of 42.3%.

Then, subtract this value from 100 to give us the final answer of 57.2%.
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Answer:

B or $12.50

Step-by-step explanation:

To get the answer, you have to solve x from the equation 4.5x = 56.25

To solve, you have to divide 4.5 from each side, making the equation look like,

x = 12.5

Hope this helps!!

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During it's first 50 days of growth a sunflower grows about 4 cm a day. Using this rate how many days will it take for the sunfl
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3 years ago
A glass is 4/7 full. Then 70 cm³ orange juice is poured in. The glass is now 3/4 full.
Nataly_w [17]

Answer:

The volume of the glass is 217.8 cm³

Step-by-step explanation:

If the glass were initially 4/7 full, that means 3/7 of the volume is still available to hold more juice.

Let v represent the volume of the glass.  

Then (3/7)v + 70 cm³ = (3/4)v.

We need to solve this for v.

Here the LCD is 28.  Thus,

(3/7)v + 70 cm³ = (3/4)v →  (12/28)v + 70 cm³ = (21/28)v.

Subtracting  (12/28)v from both sides, we get:

70 cm³ = (9/28)v.

We can isolate v by mult. both sides by the inverse of 9/28, which is 28/9:

(28/9)(70 cm³) = v

The volume of the glass is 217.8 cm³

3 0
3 years ago
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3 years ago
Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29). Round your answer to at least three deci
ss7ja [257]

Answer:

a) 0.76197086

b) -1.73406361

Step-by-step explanation:

a)

Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 < t < 1.29)

P(-1.29 < t < 1.29) would be the area under the t distribution curve with 7 degrees of freedom between -1.29 and 1.29, that is in the interval (-1.29, 1.29).

This can be done the old style by looking up in a table or by using the technology with a spreadsheet.

In Excel, the function TDIST(x,n,2) with x>0 gives the area outside the interval (-x, x) of the t distribution with n degrees of freedom.

So TDIST(1.29,7,2) gives the area outside (-1.29, 1.29).

If we subtract this value from 1 we get the desired result

Hence  

P(-1.29 < t < 1.29) = 1 - TDIST(1.29,7,2) = 1 - 0.23802914 = 0.76197086

In OpenOffice Calc, the function is the same replacing “,” with “;”  

That is

P(-1.29 < t < 1.29) = 1 - TDIST(1.29;7;2) = 0.76197086

b)

Consider a t distribution with 18 degrees of freedom. Find the value of c such that P(t≤ c) = 0.05

We are looking for a point c such that the area of the t distribution with 18 degrees of freedom to the left of c is 0.05

In Excel, the inverse function of TDIST is TINV.  

TINV(p*2,n) with p>0 gives the point c such that the area of the t distribution with n degrees of freedom to the right of c is p.  

Since <em>the t distribution is symmetric with respect to 0</em>, -c would be a point such that the area to the left of -c is p.

So we want to compute  in Excel

-TINV(0.05*2,18) = -1.73406361

In OpenOffice Calc  

-TINV(0.05*2;18) = -1.73406361

3 0
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