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3 years ago

What are the first 4 terms in the sequence represented by the expression 2n+3

Mathematics

2 answers:

torisob [31]3 years ago

5 0

The answer would be A

otez555 [7]3 years ago

5 0

2n + 3

n = 1, 2, 3, 4,...

For n = 1, 2n + 3 = 2*1 + 3 = 2 + 3 = 5

For n = 2, 2n + 3 = 2*2 + 3 = 4 + 3 = 7

For n = 3, 2n + 3 = 2*3 + 3 = 6 + 3 = 9

For n = 4, 2n + 3 = 2*4 + 3 = 8 + 3 = 11

So the first four terms are 5, 7, 9, 11. So it is option C. And yes you are right.

How did you manage to get it right even with a foggy head.

Hope this helps.

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I need help.. with thissss

damaskus [11]

9514 1404 393

Answer:

7.06

Step-by-step explanation:

This triangle can be solved a couple of ways. In the end, they amount to the same thing.

1) The area is ...

A = 1/2bh = 1/2(8)(15) = 60 . . . using DG as the base

Using GE as the base, the height (DF) is ...

A = (1/2)(17)(DF)

2(60)/17 = DF = 120/17

DF ≈ 7.06

2) Using similar triangles, we can find the ratio of the long side to the hypotenuse as ...

(long side)/(hypotenuse) = DE/GE = DF/DG

DF = DG(DE/GE) = 8(15/17) = 120/17

DF ≈ 7.06

5 0

2 years ago

Write the first five terms of the arithmetic sequence. a1 = 9; d = -2

asambeis [7]

Answer:

nth term = 1 1/2n -1

Step-by-step explanation:

The arithmetic sequence formula is:

a n = a 1 + ( n – 1 ) d

Where: a n

is the nth term in the sequence

a 1

is the first term in the sequence

is the term you are solving for

is the common difference for any pair of consecutive numbers in the sequence.

First Term or

n = 1 :

This is given in the problem.

a 1 = 9

Second Term or

n = 2 :

Substitute

2 for n

in the formula and substitute the values from the problem giving:

a 2 = 9 + ( ( 2 – 1 ) × -2 )

a 2 = 9 + ( 1 × -2 )

a 2 = 9 +-2

a 2 = 7

Fifth Term or n = 5 :

Substitute in the formula and substitute the values from the problem giving:

a 5 = 9+ ( ( 5– 1 ) × -2 )

a 5 = 9 + ( 4 × -2 )

a 2 = 9 + -8

a 2 = 1

Using this same process you should be able to determiner the

Third Term or n = 3

: and Fourth Term or n = 4 :

4 0

2 years ago

The joint F.D.P of a bivariate VA (X, Y) is given

Nat2105 [25]

Answer:

Step-by-step explanation:

8 0

3 years ago

Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people

Leona [35]

Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

$p = 0.85, n = 6$ .

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

$P(X < 6) + P(X = 6) = 1$

$P(X < 6) = 1 - P(X = 6)$

In which:

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

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229$

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X \leq 4) = 1 - P(X > 4)$

In which

$P(X > 4) = P(X = 5) + P(X = 6)$

$P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993$

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764$

$P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236$

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

$P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960$

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

$P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376$

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

5 0

3 years ago

Mrs.Kennedy is buying pencils for each of 315 students at Hamilton Elementary. The pencils are sold in boxes of tens. How can sh

g100num [7]

Mrs Kennedy is buying pencils for 315 students.

The pencils are sold in boxes of 10.

Therefore the number of pencils she needs to buy should be a multiple of 10, so that she can buy boxes of 10 pencils each.

number of students are 315 but 315 is not a multiple of 10

so we have to round off 315 to the nearest ten, that's 320.

so then she has to buy 320 pencils

number of boxes she needs - 320 / 10 = 32 boxes of pencils

she only needs pencils for 315 students so she will have 5 extra pencils

she will have to buy 32 boxes

4 0

3 years ago

Read 2 more answers