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enot [183]
3 years ago
6

What is 12.8×3/4 pls answer this asap​

Mathematics
2 answers:
patriot [66]3 years ago
4 0

Answer:

it would be 9.6

Step-by-step explanation:

natima [27]3 years ago
3 0

Answer:

9.6

Step by step explanation:

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Jordi trains on his bicycle by riding for a total of 300 miles each week.
Alex73 [517]
Since y represents miles and x represents weeks, it would be y=300x
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Can some one help<br> The last two are <br> H and J
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the correct answer for this problem is J

5 0
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What is the median of the data set below? 57, 21, 11, 35, 43, 19​
Naily [24]

Answer:

28

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
For a normal distribution with μ=500 and σ=100, what is the minimum score necessary to be in the top 60% of the distribution?
STatiana [176]

Answer:

475.

Step-by-step explanation:

We have been given that for a normal distribution with μ=500 and σ=100. We are asked to find the minimum score that is necessary to be in the top 60% of the distribution.

We will use z-score formula and normal distribution table to solve our given problem.

z=\frac{x-\mu}{\sigma}

Top 60% means greater than 40%.

Let us find z-score corresponding to normal score to 40% or 0.40.

Using normal distribution table, we got a z-score of -0.25.

Upon substituting our given values in z-score formula, we will get:

-0.25=\frac{x-500}{100}

-0.25*100=\frac{x-500}{100}*100

-25=x-500

-25+500=x-500+500

x=475

Therefore, the minimum score necessary to be in the top 60% of the distribution is 475.

3 0
3 years ago
Lin runs 5 laps around a track in 8 minutes.
andrezito [222]

Answer:

To run one lap Lin takes 1.6 minutes.

Step-by-step explanation:

To find this answer you would do 8 divided by 5. Always remember when dividing time goes first unless the instructions / directions / your teacher says differently. So, when you do 8 divided by 5 the answer you get is 1.6, and there is your answer. Also to check your answer you can do 1.6 times 5, and you get 8, so you know that 1.6 is the correct answer.

5 0
3 years ago
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