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2 years ago

Find the 4th term in the following sequence (n = 4)

1 answer:

lisabon 2012 [21]2 years ago

5 0

this formula, doesn't rely on a product, relies on a "sum", or is namely an arithmetic sequence, aₙ₋₁ -3 is another way of saying, the value of the previous term minus 3, so it relies on the ordinal value of a term, so is an recursive formula, well, let's get it when n = 4.

$\begin{array}{ccll} order&term&value\\ \cline{1-3} 1&a_1&8\\ 2&a_2&a_{2-1}-3\\ &&a_1-3\\ &&8-3\\ &&5\\ 3&a_3&a_{3-1}-3\\ &&a_2-3\\ &&2\\ 4&a_4&a_{4-1}-3\\ &&a_3-3\\ &&2-3\\ &&-1 \end{array}$

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Please help me! I will mark brainliest! Thanks!

Andreas93 [3]

Answer: Group 12

Step-by-step explanation: The element is Mercury.

8 0

3 years ago

Read 2 more answers

Solve for y math problem

o-na [289]

Answer:

The answer is 4/5

Step-by-step explanation:

-1/2 y ≤ -2/5

1/2y ≤ 2/5

y ≤ 2 × 2/5

y ≤ 4/5

Thus, The value of y is 4/5

<u>-TheUnknownScientist 72</u>

6 0

2 years ago

A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed

tresset_1 [31]

Answer:

a) $z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) $P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

c) $z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

$X \sim N(\mu = 34, \sigma=2.5)$

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

$P(34 < X$

We can find the number of deviation from the mean with the z score formula:

$z= \frac{X -\mu}{\sigma}$

And replacing we got

$z= \frac{34-34}{2.5}= 0$

$z= \frac{39-34}{2.5}= 2$

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

$P(X$

$z= \frac{31.5-34}{2.5}= -1$

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

$P(29 < X$

And replacing we got:

$z= \frac{29-34}{2.5}= -2$

$z= \frac{36.5-34}{2.5}= 1$

7 0

3 years ago

Determine whether the three segment lengths will produce a triangle. Type yes or no in the space provided.

tresset_1 [31]

Answer:

20 , 20 , 30 will produce a triangle

Step-by-step explanation:

* Lets study how to know if the lengths of the three segments

can formed a triangle

- It is a fact that in any triangle the sum of the smallest two sides

must be greater than the largest side

- Lets study some examples

# If the lengths of the three segments are 5 , 6 , 7

∵ 5 and 6 are the smallest

∴ 5 + 6 = 11

∵ 11 > 7 ⇒ the sum greater than the 3rd side

∴ 5 , 6 , 7 can formed a triangle

# If the lengths of the three segments are 5 , 7 , 12

∵ 5 and 7 are the smallest

∴ 5 + 7 = 12

∵ 12 = 12 ⇒ the sum equal the 3rd side

∴ 5 , 7 , 12 can not formed a triangle

# If the lengths of the three segments are 10 , 12 , 24

∵ 10 and 12 are the smallest

∴ 10 + 12 = 22

∵ 22 < 24 ⇒ the sum less than the 3rd side

∴ 10 , 12 , 24 can not formed a triangle

* Now lets solve the problem

∵ The length of the three segments are 20 , 20 , 30

∵ 20 and 20 are the smallest

∴ 20 + 20 = 40

∵ 40 > 30 ⇒ the sum greater than the 3rd side

∴ 20 , 20 , 30 will produce a triangle

5 0

3 years ago

a globe has a diameter of 12 inches. it fits inside a cube-shaped box that has a side length of 12 inches. What is the volume, r

baherus [9]

Answer:

823.68 inches

Step-by-step explanation:

7 0

3 years ago