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2 years ago

15

There is a bag of marbles that contains 4 blue marbles, 3 red marbles, 2 green marbles, and 1 brown marble. If one was picked at

random, not replaced, and another was picked, what is the probability of both marbles selected to be blue?

1 answer:

Semenov [28]2 years ago

7 0

Answer:

2/15

Step-by-step explanation:

P(B,B) = 4/10 x 3/9

= 2/5 x 1/3 or 2/15

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lorasvet [3.4K]

Answer:

the answer is 40 because you count all the squares that are in the shape

Step-by-step explanation:

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3 years ago

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One rule of thumb for estimating crowds is that each person occupies 2.3 square feet. Use this value to estimate the size of a c

dybincka [34]

Answer:

150,595

Step-by-step explanation:

First, convert miles to feet:

4.1 mi × (5280 ft/mi) = 21648 ft

The area that the crowd takes up is:

2 × 8 ft × 21648 ft = 346368 ft²

The number of people is:

346368 ft² × (1 person / 2.3 ft²) = 150594.8

Rounding to the nearest integer, there are approximately 150,595 people in the crowd.

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2 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

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

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3 years ago

Serena was planning out her model railroad setup, and wants to be able to boast about it on her social media. She wants the cent

DerKrebs [107]

Answer:4

Step-by-step explanation:

3 0

3 years ago

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Evaluate <br> 3^-3*3^4*3*3^-5

miss Akunina [59]

To multiply exponents with like bases, simply add the exponents...

3^-3 × 3^4 × 3 × 3^-5 =

add exponents across, remember just 3 means 3^1.

-3 + 4 + 1 + -5 = -3

answer is 3^-3

8 0

3 years ago