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Art [367]
3 years ago
9

Which set could represent the side lengths of a triangle?

Mathematics
2 answers:
Doss [256]3 years ago
6 0

So, the sum of two sides have to be greater than the third side for it to create a triangle.

Option A:

4 + 6 = 10

The first 2 sides aren't greater than the third side so, this is incorrect.

Option B:

4 + 5 = 9

The first 2 sides aren't greater than the third side so, this incorrect.

Option C:

5 + 6 = 11

The first 2 sides aren't greater than the third side so, this incorrect.

Option D:

5 + 5 = 10

The first 2 sides are greater than the third side so, this correct.

Therefore, the answer is Option D.

Best of Luck!

iogann1982 [59]3 years ago
5 0
The last one is correct
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Five million bacteria live in an organism. The doubling period is 1.5 hours. How many will there be after:
finlep [7]
We write the generic exponential equation:
 y = A (b) ^ t
 Where,
 A: initial population
 B: growth rate
 t: time in hours.
 Substituting values we have:
 y = 5 (2) ^ ((1 / 1.5) * t)
 For t = 9 hours we have:
 y = 5 (2) ^ ((1 / 1.5) * 9)
 y = 320 million
 For t = 24 hours (one day) we have:
 y = 5 (2) ^ ((1 / 1.5) * 24)
 y = 327680 million
 For t = 2 hours we have:
 y = 5 (2) ^ ((1 / 1.5) * 2)
 y = 12.5992105 million
 Answer:
 
a) t hours?
 
y = 5 (2) ^ ((1 / 1.5) * t)
 
b) 9 hours?
 
y = 320 million
 
c) 1 day?
 
y = 327680 million
 
d) 2 hours?
 
y = 12.5992105 million
8 0
3 years ago
Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let AAA be the event that the six-sided d
sammy [17]

The question is incomplete. Here is the complete question.

Suppose that Adam rolls a fair six-sided die and a fair four-sided die simultaneously. Let A be the event that the six-sided die is an even number and B be the event that the four-sided die is an odd number. Using the sample space of possible outcomes below, answer each of the following questions.

What is P(A), the probabillity that the six-sided die is an even number?

What is P(B), the probability of the four-sided die is an odd number?

What is P(A and B), the probability that the six-sided die is an even number  <em>and</em> the four-sided die is an odd number?

Are events A and B independent?

a) Yes, events A and B are independent events.

b) No, events A and B are not independent events.

Answer: P(A) = 1/2

P(B) = 1/2

P(A and B) = 1/4

a) Yes, events A and B are independent events.

Step-by-step explanation: The event A is related to a six-sided die, so total possibilities is 6.

For a six-sided die to show a even number, there are 3 possibilities: (2,4,6)

so, P(A) = 3/6 = 1/2

The event B is for a 4-sided die, i.e. total possibilities is 4.

To show an odd number, there are 2 possibilities: (1,3).

Then, P(B) = 2/4 = 1/2

Now, the probability of occuring A and B is:

P(A and B) = P(A).P(B)

P(A and B) = 1/2*1/2

P(A and B) = 1/4

The events are <u>independent</u> events because the probability of A happening does not influence the occuring of event B.

5 0
3 years ago
A softball coach has ordered softballs for two different leagues. The Junior League uses
dedylja [7]

The number of 11-inch softball is 70 and the number of 12-inch softball is 50.

<u>Step-by-step explanation</u>:

<u>Given that,</u>

  • The cost of 11-inch softball = $2.50
  • The cost of 12-inch softball = $3.50

<u>Let us assume,</u>

  • The number of 11-inch softball be 'x'.
  • The number of 12-inch softball be 'y'.

<u>Forming the equation to solve x and y values :</u>

  • The total number of softball ordered = 120
  • The total cost for 120 softballs = $350

x + y = 120  -------(1)

2.5x + 3.5y = 350   --------(2)

<u>Multiply eq(1) by 2.5 and subtract eq(2) from eq(1)</u>,

 2.5x +2.5y = 300

-<u>(2.5x +3.5y = 350)</u>

  <u>        -1y     =  -50</u>

Therefore the value of y = 50.

The number of 12-inch softball is 50.

<u>Substitute y=50 in eq(1),</u>

x+50 = 120

x = 120-50

x = 70

The number of 11-inch softball is 70.

5 0
3 years ago
Click to correct any capitalization errors.
Naddika [18.5K]

Answer:

Step-by-step explanation:

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5 0
3 years ago
A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa
UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so \mu = \frac{14}{60} = 0.2333

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0
4 years ago
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