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3 years ago

Find the 13th term of the sequence 5, 8, 11, 14, 17...

Mathematics

2 answers:

snow_tiger [21]3 years ago

6 0

5,8,11,14,17,20,23,26,29,32,35,38,41

Add each number by 3

13th term is 41 ( first choice)

MrRissso [65]3 years ago

4 0

Answer:41

Step-by-step explanation:

The numbers go up by 3

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Evaluate expression when a=6 a2-6a+4

Mekhanik [1.2K]

The first step to solving this problem is to plug in 6 where you see a in the expression
-6² - 6 (-6) + 4
the first step to solving this is to evaluate the power
-36 - 6x (-6) + 4
next you will multiply the numbers
-36 + 36 + 4
finally,, eliminate the opposite of the expression to find your answer
4
this means that the correct answer to your question is 4.
let me know if you have any further questions
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7 0

4 years ago

Read 2 more answers

A sack of potatoes weighs 14 pounds 9 ounces after wendy makes potato salad for a picnic the sack weights 9 pounds 14 ounces wha

dlinn [17]

For this case we have the following conversion:
1 pound = 16 ounces
Then we rewrite the sack weights:
14 pounds 9 ounces = 14 9/16
9 pounds 14 ounces = 9 14/16 = 9 7/8
Then, we subtract both values:
(14 9/16) - (9 7/8) = 4.6875 = 4 + 0.6875
Rewriting:
4 + 0.6875 = 4 + 6875/1000
4 + 1375/200 = 4 + 275/40
4 + 275/40 = 4 + 55/8
Answer:
the weight of the potatoes wendy use for the potato salad is:
4 + 55/8 pounds

5 0

4 years ago

A recent study suggested that 70% of all eligible voters will vote in the next presidential election. Suppose 20 eligible voters

natita [175]

Answer:

0.0479 = 4.79% probability that fewer than 11 of them will vote

Step-by-step explanation:

For each voter, there are only two possible outcomes. Either they will vote, or they will not. The probability of a voter voting is independent of any other voter, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

70% of all eligible voters will vote in the next presidential election.

This means that $p = 0.7$

20 eligible voters were randomly selected from the population of all eligible voters.

This means that $n = 20$

What is the probability that fewer than 11 of them will vote?

This is:

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{20,10}.(0.7)^{10}.(0.3)^{10} = 0.0308$

$P(X = 9) = C_{20,9}.(0.7)^{9}.(0.3)^{11} = 0.0120$

$P(X = 8) = C_{20,8}.(0.7)^{8}.(0.3)^{12} = 0.0039$

$P(X = 7) = C_{20,7}.(0.7)^{7}.(0.3)^{13} = 0.0010$

$P(X = 6) = C_{20,10}.(0.7)^{6}.(0.3)^{12} = 0.0002$

$P(X = 5) = C_{20,5}.(0.7)^{5}.(0.3)^{15} \approx 0$

The probability of 5 or less voting is very close to 0, so they will not affect the outcome. Then

$P(X < 11) = P(X = 10) + P(X = 9) + P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4) + P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0) = 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.0479$