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Dahasolnce [82]
3 years ago
11

Graph the linear inequality X<1

Mathematics
1 answer:
tester [92]3 years ago
8 0
Ohh wow nice, do your own work lazy person
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Simplify. 10y(3x – 7z) <br><br> 30xy – 70yz <br> 103xy – 107yz <br> 30xy – 7z <br> 3x – 70yz
oksian1 [2.3K]

30yx−70yz is your answer

6 0
3 years ago
Read 2 more answers
Can someone help me with this question: What is the 40th term of these sequences below. 13, 26, 39, 52,...... 6, 12, 18, 24,....
Nina [5.8K]

Answer:520; 240

Step-by-step explanation:

a. 13, 26, 39, 52,......

a = First term = 13

d = common difference = 26 - 13 = 13

40th term = a + (n - 1)d = a + (40-1)d = a + 39d

= 13 + (39 × 13)

= 13 + 507

= 520

b. 6, 12, 18, 24,.......

a = First term = 6

d = common difference = 12 - 6 = 6

40th term = a + 39d

= 6 + 39(6)

= 6 + 234.

= 240

8 0
3 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
Simplify: 4.1c-(3.2c)-0.1c
OverLord2011 [107]

Answer:

0.8c

Step-by-step explanation:

4 0
3 years ago
General Electric Company has an assembly line where light bulbs are produced. The ratio of defective bulbs to good bulbs produce
kati45 [8]

Answer: 200 bulbs will not be defective.

Step-by-step explanation:

The ratio of defective bulbs to good bulbs produced each day is 2 to 10. This ratio can also be expressed as 1 to 5 by reducing to lowest terms.

The total ratio is the sum of the proportions.

Total ratio = 1 + 5 = 6

This means that if n bulbs is produced, the number of defective bulbs would be

1/6 × n

The number of non defective would be

5/6 × n

Since n = 240, then the number of bulbs that will not be defective is

5/6 × 240 = 200 bulbs

4 0
3 years ago
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