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3 years ago

How to write 9 and 9 thousandths in standard form

Mathematics

1 answer:

Paha777 [63]3 years ago

4 0

Answer:

9.009

Step-by-step explanation:

As a mixed number, it is 9 9/1000.

The "thousandths" place in a decimal number is the third digit to the right of the decimal point, so a 9 in that place signifies 9 thousandths, or 9/1000. Adding 9 units gives ...

9 9/1000 = 9 + 0.009 = 9.009

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Answer:

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Step-by-step explanation:

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2 years ago

What is 18/32 in simplest form

JulsSmile [24]

Answer:

Step-by-step explanation:

18/32 is the same as 9/16 or 0.5625

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2 years ago

How to factor polynomials

Ierofanga [76]

Check for the GCF first. ...Multiply the quadratic term and the constant term. ...Write down all the factors of the result, in pairs. ...From this list, find the pair that adds to produce the coefficient of the linear term.

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3 years ago

Two terms of a geometric sequence are given. Find the first five terms. Please help asap

Zepler [3.9K]

Answer:

4, 8, 16, 32, 64

Step-by-step explanation:

The nth term of a geometric sequence is

$a_{n}$ = a₁ $(r)^{n-1}$

Given

a₇ = 256 and a₁₀ = 2048 , then

a₁ $r^{6}$ = 256 → (1)

a₁ $r^{9}$ = 2048 → (2)

Divide (2) by (1)

$\frac{a_{1}r^{9} }{a_{1}r^{6} }$ = $\frac{2048}{256}$

r³ = 8 ( take the cube root of both sides )

r = $\sqrt[3]{8}$ = 2

Substitute r = 2 into (1)

a₁ × $2^{6}$ = 256

a₁ × 64 = 256 ( divide both sides by 64 )

a₁ = 4

Then

a₁ = 4

a₂ = 2a₁ = 2 × 4 = 8

a₃ = 2a₂ = 2 × 8 = 16

a₄ = 2a₃ = 2 × 16 = 32

a₅ = 2a₄ = 2 × 32 = 64

7 0

2 years ago

A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas Cit

Orlov [11]

Using the binomial distribution, we have that:

a) 0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

b) 0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

c) The expected number of people is 4, with a variance of 20.

For each person, there are only two possible outcomes. Either they attended a game, or they did not. The probability of a person attending a game is independent of any other person, which means that the binomial distribution is used.

Binomial probability distribution

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

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

The parameters are:

p is the probability of a success on a single trial.

n is the number of trials.

p is the probability of a success on a single trial.

The expected number of trials before q successes is given by:

$E = \frac{q(1-p)}{p}$

The variance is:

$V = \frac{q(1-p)}{p^2}$

In this problem, 0.2 probability of a finding a person who attended the last football game, thus $p = 0.2$ .

Item a:

None of the first three attended, which is P(X = 0) when n = 3.
Fourth attended, with 0.2 probability.

Thus:

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

$P(X = 0) = C_{3,0}.(0.2)^{0}.(0.8)^{3} = 0.512$

$0.2(0.512) = 0.1024$

0.1024 = 10.24% probability that the marketing representative must select 4 people to find one who attended the last home football game.

Item b:

This is the probability that none of the first six went, which is P(X = 0) when n = 6.

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

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

0.2621 = 26.21% probability that the marketing representative must select more than 6 people to find one who attended the last home football game.

Item c:

One person, thus $q = 1$ .

The expected value is:

$E = \frac{q(1-p)}{p} = \frac{0.8}{0.2} = 4$

The variance is:

$V = \frac{0.8}{0.04} = 20$

The expected number of people is 4, with a variance of 20.

A similar problem is given at brainly.com/question/24756209

3 0

1 year ago