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2 years ago

Jessie says to round 763,400 to the nearest ten thousand he will round to 770,000 Is he right ?

Mathematics

2 answers:

velikii [3]2 years ago

6 0

Yes, he is correct. because if it's 750,000 or more, than you round up. otherwise, you round down.

Ludmilka [50]2 years ago

4 0

Yes because 5 or more rounds up but 4 or less rounds down.

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Draw apicture to show how many hundreds there are in 1 thousand

Delicious77 [7]

So that answer is in this picture that I wrote on my white board, count all of the hundreds I drew on the board.

6 0

3 years ago

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How do you factor 3x^2-7x-2

vivado [14]

3x^2-7x-2

doesn't factor

You can not factor this

I hope that's help !

3 0

3 years ago

Under his cell phone plan, Owen pays a flat cost of $67.50 per month and $4 per gigabyte. He wants to keep his bill at $71.90 pe

Agata [3.3K]

Answer:

1.1 gigabytes

Step-by-step explanation:

Let us represent:

The number of gigabytes = g

Under his cell phone plan, Owen pays a flat cost of $67.50 per month and $4 per gigabyte. He wants to keep his bill at $71.90 per month.

The Equation is given as:

$71.90 = $67.50 + $4 × g

71.90 = 67.50 + 4g

71.90 - 67.50 = 4g

4.4 = 4g

x = 4.4/4

x = 1.1 gigabytes

Therefore, the number of gigabytes of data Owen can use while staying within his budget is 1.1 gigabytes

8 0

2 years ago

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if you roll a fair 6-sided die 9 times, what is the probability that at least 2 of the rolls come up as a 3 or a 4?

Kay [80]

Using the binomial distribution, it is found that there is a 0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For each die, there are only two possible outcomes, either a 3 or a 4 is rolled, or it is not. The result of a roll is independent of any other roll, hence, the binomial distribution is used to solve this question.

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:

x is the number of successes.

n is the number of trials.

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

In this problem:

There are 9 rolls, hence $n = 9$ .

Of the six sides, 2 are 3 or 4, hence $p = \frac{2}{6} = 0.3333$

The desired probability is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1)$

Then

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

$P(X = 0) = C_{9,0}.(0.3333)^{0}.(0.6667)^{9} = 0.026$

$P(X = 1) = C_{9,1}.(0.3333)^{1}.(0.6667)^{8} = 0.117$

Then:

$P(X < 2) = P(X = 0) + P(X = 1) = 0.026 + 0.117 = 0.143$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.143 = 0.857$

0.857 = 85.7% probability that at least 2 of the rolls come up as a 3 or a 4.

For more on the binomial distribution, you can check brainly.com/question/24863377

7 0

2 years ago

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True [87]

Answer:

The correct answer to this question is complementars

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3 years ago