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Paraphin [41]
3 years ago
11

What is the exponent of 10? (Thank you for helping!)

Mathematics
1 answer:
Lemur [1.5K]3 years ago
8 0

Answer:

exponent is 0

Step-by-step explanation:

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Answer for Brainliest<br><br> 1)7/4+5/8=<br> 2)3/2+4/3=<br> 3)2/3+10/7=<br> 4)4-3/17
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Answer:

See below.

Step-by-step explanation:

1) 7/4 + 5/8

= 14/8 + 5/8

= 19/8 or 2 3/8

2) 3/2 + 4/3

= 9/6 + 8/6

= 15/6

= 5/2 or 2 1/2

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= 14/21 + 30/21

= 44/21 or 2 2/21

4) 4 - 3/17

= 68/17 - 3/17

= 65/17 or 3 14/17

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Step-by-step explanation:

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A random sample is drawn from a normally distributed population with mean μ = 31 and standard deviation σ = 1.9. Calculate the p
lesya692 [45]

Answer:

For sample size n = 39 ; P(X < 31.6) = 0.9756

For sample size n = 76 ; P(X < 31.6) = 0.9970

Step-by-step explanation:

Given that:

population mean μ = 31

standard deviation σ = 1.9

sample mean  \overline X = 31.6

Sample size n                 Probability

39

76

The probabilities that the sample mean is less than 31.6 for both sample size can be computed as  follows:

For sample size n = 39

P(X < 31.6) = P(\dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}})

P(X < 31.6) = P(\dfrac{31.6 - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{39}}})

P(X < 31.6) = P(Z< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{39}}})

P(X < 31.6) = P(Z< \dfrac{0.6}{\dfrac{1.9 }{6.245}})

P(X < 31.6) = P(Z< 1.972)

From standard normal  tables

P(X < 31.6) = 0.9756

For sample size n = 76

P(X < 31.6) = P(\dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{\overline X - \mu}{\dfrac{\sigma }{\sqrt{n}}})

P(X < 31.6) = P(\dfrac{31.6 - \mu}{\dfrac{\sigma }{\sqrt{n}}}< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{76}}})

P(X < 31.6) = P(Z< \dfrac{31.6 - 31}{\dfrac{1.9 }{\sqrt{76}}})

P(X < 31.6) = P(Z< \dfrac{0.6}{\dfrac{1.9 }{8.718}})

P(X < 31.6) = P(Z< 2.75)

From standard normal  tables

P(X < 31.6) = 0.9970

6 0
3 years ago
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