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Verizon [17]
3 years ago
8

there are 175 total balloons the ratio of gold to silver balloons is 3:2 how many balloons are gold out of the 175 balloons

Mathematics
1 answer:
egoroff_w [7]3 years ago
5 0
175/3 = 117 gold 58 silver
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Please help me I don't understand on how to put these numbers on the number line
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Answer: I had to anotate it , so you have to download that file. Good luck with what you doing.

Step-by-step explanation:

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7 0
3 years ago
Find x. A. 21√2 B. 7 C. 21√3 over 2 D. 21√2 over 2
kkurt [141]

Answer:

D

Step-by-step explanation:

<em>for </em><em>you </em><em>to </em><em>find </em><em>x </em><em>you </em><em>first </em><em>have </em><em>to </em><em>find </em><em>the </em><em>adjacent</em><em> </em><em>of </em><em>the </em><em>4</em><em>5</em><em>°</em><em> </em><em>angle </em><em>you </em><em>can </em><em>do </em><em>that </em><em>by </em><em>using</em><em> </em><em>the </em><em>other </em><em>triangle.</em><em>u</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>the </em><em>sin </em><em>ratio</em>

<em>sin60=</em><em>opposite</em><em>/</em><em>hypotenuse</em>

<em>sin60=</em><em>a/</em><em>7</em><em>√</em><em>3</em>

<em>a</em><em>=</em><em>1</em><em>0</em><em>.</em><em>5</em>

<em>then </em><em>after </em><em>you </em><em>have </em><em>found</em><em> </em><em>the </em><em>adjacent</em><em> </em><em>you </em><em>can </em><em>use </em><em>the </em><em>cos </em><em>ratio</em>

<em>cos45=</em><em>adjacent/</em><em>hypotenuse</em>

<em>cos45=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>/</em><em>x</em>

<em>cos45x/</em><em>cos45=</em><em>1</em><em>0</em><em>.</em><em>5</em><em>/</em><em>cos45</em>

<em>x=</em><em>1</em><em>4</em><em>.</em><em>8</em><em>4</em><em>9</em>

<em>which </em><em>is </em><em>the </em><em>same </em><em>as </em><em>2</em><em>1</em><em>√</em><em>2</em><em> </em><em>over </em><em>2</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

5 0
3 years ago
Read 2 more answers
I really need help!! <br> what number is equivalent to the fraction?<br> 5<br> 9
dangina [55]

Answer:

0.55555555555555555555555555556

<em>(The 5 is repeating, so I rounded the last digit to 6.)</em>

Step-by-step explanation:

Divide 5/9. You get 0.5555... and the 5 is repeating.

5 0
3 years ago
Assume that there is a 4​% rate of disk drive failure in a year. a. If all your computer data is stored on a hard disk drive wit
kap26 [50]

Answer:

a) 99.84% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

b) 99.999744% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

Step-by-step explanation:

For each disk drive, there are only two possible outcomes. Either it works, or it does not. The disks are independent. So we use the binomial probability distribution 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.

4​% rate of disk drive failure in a year.

This means that 96% work correctly, p = 0.96

a. If all your computer data is stored on a hard disk drive with a copy stored on a second hard disk​ drive, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

This is P(X \ geq 1) when n = 2

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = C_{2,0}.(0.96)^{0}.(0.04)^{2} = 0.0016

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0016 = 0.9984

99.84% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

b. If copies of all your computer data are stored on four independent hard disk​ drives, what is the probability that during a​ year, you can avoid catastrophe with at least one working​ drive?

This is P(X \ geq 1) when n = 4

We know that either none of the disks work, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

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

P(X = 0) = C_{4,0}.(0.96)^{0}.(0.04)^{4} = 0.00000256

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000256 = 0.99999744

99.999744% probability that during a​ year, you can avoid catastrophe with at least one working​ drive

7 0
3 years ago
HEO asap 8th grade math
Ierofanga [76]

a distinct section of a piece of writing, usually dealing with a single theme and indicated by a new line, indentation, or numbering.

5 0
3 years ago
Read 2 more answers
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