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2 years ago

5

Please help you will feel good and u will get good deeds

2 answers:

KIM [24]2 years ago

7 0

Answer:

5 or 6

Step-by-step explanation:

ikadub [295]2 years ago

3 0

Answer:

5

Step-by-step explanation:

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The earth is about 2.5578*10'13 miles

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2 years ago

Please help me ASAP I’ll mark Brainly

kondor19780726 [428]

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-2.3 repeating

Step-by-step explanation:

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3 years ago

generally people cover 110 yards in 11 seconds. express the speed in feet per second. (3 feet = 1 yard)

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3 years ago

The following table shows the estimated populations and annual growth rates for four countries in the year 2000. Find the

gladu [14]

Answer:

Part 1) Australia $19,751,012\ people$

Part 2) China $1,319,645,764\ people$

Part 3) Mexico $109,712,539\ people$

Part 4) Zaire $60,534,681\ people$

Step-by-step explanation:

we know that

The equation of a exponential growth function is given by

$P(t)=a(1+r)^t$

where

P(t) is the population

t is the number of years since year 2000

a is he initial value

r is the rate of change

Part 1) Australia

we have

$a=19,169,000\\r=0.6\%=0.6\100=0.006$

substitute

$P(t)=19,169,000(1+0.006)^t$

$P(t)=19,169,000(1.006)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=19,169,000(1.006)^5=19,751,012\ people$

Part 2) China

we have

$a=1,261,832,000\\r=0.9\%=0.9\100=0.009$

substitute

$P(t)=1,261,832,000(1+0.009)^t$

$P(t)=1,261,832,000(1.009)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=1,261,832,000(1.009)^5=1,319,645,764\ people$

Part 3) Mexico

we have

$a=100,350,000\\r=1.8\%=1.8\100=0.018$

substitute

$P(t)=100,350,000(1+0.018)^t$

$P(t)=100,350,000(1.018)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=100,350,000(1.018)^5=109,712,539\ people$

Part 4) Zaire

we have

$a=51,965,000\\r=3.1\%=3.1\100=0.031$

substitute

$P(t)=51,965,000(1+0.031)^t$

$P(t)=51,965,000(1.031)^t$

Find the expected population in 2025,

Find the value of t

t=2005-2000=5 years

substitute the value of t in the equation

$P(5)=51,965,000(1.031)^5=60,534,681\ people$

8 0

3 years ago

An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability tha

TEA [102]

Answer:

The probability is $P(X >300 ) = 0.97219$

Step-by-step explanation:

From the question we are told that

The capacity of an Airliner is k = 300 passengers

The sample size n = 320 passengers

The probability the a randomly selected passenger shows up on to the airport

$p = 0.96$

Generally the mean is mathematically represented as

$\mu = n* p$

=> $\mu = 320 * 0.96$

=> $\mu = 307.2$

Generally the standard deviation is

$\sigma = \sqrt{n * p * (1 -p ) }$

=> $\sigma = \sqrt{320 * 0.96 * (1 -0.96 ) }$

=> $\sigma =3.50$

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

$P(X > k ) = P( \frac{ X -\mu }{\sigma } > \frac{k - \mu}{\sigma } )$

Here $\frac{ X -\mu }{\sigma } =Z (The \ standardized \ value \ of \ X )$

=> $P(X >300 ) = P(Z > \frac{300 - 307.2}{3.50} )$

Now applying continuity correction we have

$P(X >300 ) = P(Z > \frac{[300+0.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )$

=> $P(X >300 ) = P(Z > -1.914 )$

From the z-table

$P(Z > -1.914 ) = 0.97219$

So

$P(X >300 ) = 0.97219$

8 0

2 years ago