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zubka84 [21]
3 years ago
8

Help pls reallyyyyyyyyyyyyyyyyyy

Mathematics
1 answer:
Ivan3 years ago
5 0

Answer:

10

Step-by-step explanation:

j=6

k=4

Rewrite the equation. So let's do that-

2.5(6*4)/6

6 and the other 6 cancel out so we are left with 2.5*4

Which is then equal to 10.

By the way, whenever you see the same number in a division problem on the numerator and denominator, just cancel them out because if you still did 2.5(6*4)/6, you would still get 10. I was just simplifying it!

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worty [1.4K]
35,000 ft - 6,000 ft = 29,000 ft
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(a) How many tens are in 457? How many whole tens?
PolarNik [594]

Answer:

  • 45,7 tens
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Step-by-step explanation:

457 / 10 = 45,7

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2 years ago
I need it ASAP, Questions are attached as a photo
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F(x) = -3 +3
Mila [183]

Answer: Y

Step-by-step explanation:

Since f(x) passes through (0,3), its inverse should pass through (3,0).

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3 0
2 years ago
he life of light bulbs is distributed normally. The variance of the lifetime is 625 and the mean lifetime of a bulb is 540 hours
almond37 [142]

Using the normal distribution, it is found that there is a 0.877 = 87.7% probability of a bulb lasting for at most 569 hours.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 540, \sigma = \sqrt{625} = 25

The probability of a bulb lasting for at most 569 hours is the <u>p-value of Z when X = 569</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{569 - 540}{25}

Z = 1.16

Z = 1.16 has a p-value of 0.877.

0.877 = 87.7% probability of a bulb lasting for at most 569 hours.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

8 0
2 years ago
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