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Deffense [45]
3 years ago
13

Please help

Mathematics
1 answer:
MrRa [10]3 years ago
4 0

Answer:

About 99.7% of births would be expected to occur within 51 days of the mean pregnancy​ length

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Standard deviation = 17.

About what percentage of births would be expected to occur within 51 days of the mean pregnancy​ length?

51/17 = 3.

So, within 3 standard deviations of the mean.

About 99.7% of births would be expected to occur within 51 days of the mean pregnancy​ length

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An account grows at an annual interest rate of r⁡​ ⁣, so it grows by a factor of x=1+r⁡​ ⁣⁣ ​⁡ each year. The function A(x)=800x
Paladinen [302]

Given:

Annual interest rate = r⁡​%

Growth factor : x = 1 + r⁡​

The below function gives the amount in the account after 4 years when the growth factor is x⁡​ ⁣⁣.

A(x)=800x^4+350x^3+500x^2+600x

To find:

The total amount in the account if the interest rate for the account is 3% each year and initial amount.

Solution:

Rate of interest = 3% = 0.03

Growth factor : x = 1 + ⁡0.03 = 1.03

We have,

A(x)=800x^4+350x^3+500x^2+600x

Substitute x=1.03 in the given function, to find the total amount in the account if the interest rate for the account is 3% each year.

A(1.03)=800(1.03)^4+350(1.03)^3+500(1.03)^2+600(1.03)

A(1.03)=800(1.12550881)+350(1.092727)+500(1.0609
)+618

A(1.03)=900.407048+382.45445+530.45+618

A(1.03)=2431.311498


A(1.03)\approx 2431.31


Therefore, the total amount in the account is 2431.31 if the interest rate for the account is 3% each year.

For initial amount the rate of interest is 0.

Growth factor : x = 1 + ⁡0 = 1

Substitute x=0 in the given function to find the initial amount.

A(1)=800(1)^4+350(1)^3+500(1)^2+600(1)

A(1)=800+350+500+600

A(1)=2250

Therefore, 2250 was put into the account at the beginning.

3 0
3 years ago
There are 22 events IN ALL at an indoor track and field meet. The ratio of TRACK events to FIELD events is 8:3. How many of the
fenix001 [56]

Answer:

There are 16 track events and 6 field events

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Factor 60x−84x using the gcf
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3 years ago
The number of column inches of classified advertisements appearing on Mondays in a certain daily newspaper is normally distribut
lana66690 [7]

Answer:

The probability there will be between 280 and 360 column inches of classified advertisement

P(280≤X≤360) = P(-2≤Z≤2) = 0.9544

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean  the population = 320

The standard deviation of the Population = 20

Let 'X' be the random variable in a normal distribution

Let 'X' = 280

Z = \frac{x-mean}{S.D} = \frac{280-320}{20} = -2

Let 'X' = 360

Z = \frac{x-mean}{S.D} = \frac{360-320}{20} = 2

<u><em>Step(iii):-</em></u>

The probability there will be between 280 and 360 column inches of classified advertisement

P(280≤X≤360) = P(-2≤Z≤2)

                        = P(z≤2) -P(z≤-2)

                       = P(z≤2)+P(z≤2)

                      = 2P(z≤2)

                     = 2×0.4772     ( from normal table)

                    = 0.9544

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3 years ago
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