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Karo-lina-s [1.5K]
3 years ago
9

In a sample of 10 cards, 4 are red and 6 are blue. If 2 cards are selected at random from the sample, one at a time without repl

acement, what is the probability that both cards are not blue?
Mathematics
1 answer:
Charra [1.4K]3 years ago
3 0
The exact answer in fraction from is 2/15

The approximate answer in decimal form is 0.1333 (rounded to four decimal places)

The decimal value 0.1333 is equivalent to 13.33%

These are three ways to say the same basic answer.

===============================================

Explanation:

We have 4 red cards and 6 blue cards giving 4+6 = 10 total.

The probability of picking not blue (aka picking red) is 4/10 because there are 4 red out of 10 total.

If we don't replace the card we picked, then we have 3 red left over out of 9 total. The probability of picking red again is 3/9

Multiply those fractions
(4/10)*(3/9) = (4*3)/(10*9) = 12/90

Now reduce
12/90 = (6*2)/(6*15) = 2/15
note how the GCF 6 cancels

Using a calculator, 
2/15 = 0.1333 approximately
The '3's after the decimal point go on forever.

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(2x2 + 2x + 3) - (x2 + 2x + 1) =
g100num [7]

Answer:

\boxed{\sf D. \ x^2 + 2}

Step-by-step explanation:

\sf Simplify \ the \ following:

\sf \implies (2 {x}^{2}  + 2x + 3) - ( {x}^{2}  + 2x + 1)

\sf - ( {x}^{2}  + 2x + 1) =  -  {x}^{2}  - 2x - 1 :

\sf \implies (2 {x}^{2}  + 2x + 3) - {x}^{2}   -  2x  -  1

\sf Grouping \ like \ terms:

\sf \implies (2 {x}^{2}   -  {x}^{2}) + (2x  - 2x)+ (3    - 1)

\sf 2 {x}^{2}  -  {x}^{2}  =  {x}^{2}  :

\sf \implies {x}^{2} + (2x  - 2x)+ (3    - 1)

\sf 2x - 2x = 0 :

\sf \implies {x}^{2} + 0+ (3    - 1)

\sf 3 - 1 = 2 :

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8 0
3 years ago
the radius ???? of a sphere is expanding at a rate of 70 cm/min. The volume of a sphere is ????=43????????3 and its surface area
enyata [817]

Answer:

\frac{dV}{dt}=1120 \pi cm^3/min

Step-by-step explanation:

We are given that

Radius of sphere expanding at the rate=\frac{dr}{dt}=70 cm/min

Volume of sphere=V=\frac{4}{3}\pi r^3

Surface area of sphere=S=4\pi r^2

We have to determine the rate  at which the volume is changing with respect to time at r= 2 cm.

V=\frac{4}{3}\pi r^3

Differentiate w.r.t time

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}

Substitute the values then we get

\frac{dV}{dt}=4\pi (2)^2(70)=1120 \pi cm^3/min

Hence, the rate at which the volume of sphere is changing is given by

\frac{dV}{dt}=1120 \pi cm^3/min

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