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miskamm [114]
3 years ago
7

Write an expression to describe a rule for the sequence. Then find the 100th term in the sequence. 5, 13, 21, 29, 37, 45, … (1 p

oint)
• 8n; 800
• 8n – 3; 797
• 3n – 8; 292
• 5 + 8n; 805
Mathematics
1 answer:
swat323 years ago
6 0
The sequence is going up by 8 so it must be 8n followed by something. 5 is 3 less than 8 so the full expression will be 8n-3.
Therefore the answer is 8n-3; 797
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The number 1.3,1/1/3,1.34 in least to greatest
elixir [45]
1.3, 1 1/3, 1.34

The decimal version of 1 1/3 is 1.333 (so on)

so 1.3 is less than 1.333 and 1.333 is less than 1.34

1.3<1.333<1.34
4 0
3 years ago
Find the values of X and Y that makes these triangles congruent by the HL theorem
muminat

Answer:

C. x = 3, y = 2

Step-by-step explanation:

If both triangles are congruent by the HL Theorem, then their hypotenuse and a corresponding leg would be equal to each other.

Thus:

x + 3 = 3y (eqn. 1) => equal hypotenuse

Also,

x = y + 1 (eqn. 2) => equal legs

✔️Substitute x = y + 1 into eqn. 1 to find y.

x + 3 = 3y (eqn. 1)

(y + 1) + 3 = 3y

y + 1 + 3 = 3y

y + 4 = 3y

y + 4 - y = 3y - y

4 = 2y

Divide both sides by 2

4/2 = 2y/2

2 = y

y = 2

✔️ Substitute y = 2 into eqn. 2 to find x.

x = y + 1 (eqn. 2)

x = 2 + 1

x = 3

8 0
3 years ago
100%
Alborosie
6.75%
1 penny = 50% chance of tail
50% x 50% x 50% x 50%= 6.75%
8 0
3 years ago
How did you use similarity to find the areas and circumferences of circles? How are the radius and diameter of a circle related?
ExtremeBDS [4]

Answer:idkidk

Step-by-step explanation:

4 0
3 years ago
In a string of 12 Christmas tree light bulbs, 3 are defective. The bulbs are selected at random and tested, one at a time, until
Juliette [100K]

Using the hypergeometric distribution, it is found that there is a 0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

In this problem, the bulbs are chosen without replacement, hence the <em>hypergeometric distribution</em> is used to solve this question.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

In this problem:

  • There are 12 bulbs, hence N = 12.
  • 3 are defective, hence k = 3.

The third defective bulb is the fifth bulb if:

  • Two of the first 4 bulbs are defective, which is P(X = 2) when n = 4.
  • The fifth is defective, with probability of 1/8, as of the eight remaining bulbs, one will be defective.

Hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,12,4,3) = \frac{C_{3,2}C_{9,1}}{C_{12,4}} = 0.2182

0.2182 x 1/8 = 0.0273.

0.0273 = 2.73% probability that the third defective bulb is the fifth bulb tested.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

8 0
2 years ago
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