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Nata [24]
3 years ago
7

Which division expression could this model represent?

Mathematics
2 answers:
erma4kov [3.2K]3 years ago
7 0
I’m pretty sure that it’s 3/0.5 Hope I helped
Contact [7]3 years ago
6 0

Answer: I think it’s 3/0.5

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Every Sunday, Tamika sells pieces of homemade fudge at a local carnival. Each piece of fudge weighs 34 pound. Next Sunday, Tamik
Aloiza [94]

Answer:

hshshshshshshshsiqkZjfdudiodjfhfbbxdudijdd

5 0
3 years ago
-4|x-11|=-16 solve for x in the equation
BigorU [14]

Answer:

x = 7 and 15

Step-by-step explanation:

Divide both sides by -4...

| x - 11 | = 4

because

| -4 | = 4 and

| 4 | = 4

x - 11 = -4 and x - 11 = 4

solve for x in both equations.

x = 7 and x = 15

5 0
4 years ago
NASA is conducting an experiment to find out the fraction of people who black out at G forces greater than 6. In an earlier stud
SSSSS [86.1K]

Answer:

The large sample n = 190.44≅190

The  large  sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

<u>Step-by-step explanation</u>:

Given  population proportion was estimated to be 0.3

p = 0.3

Given maximum of error E = 0.04

we know that maximum error

M.E = \frac{Z_{\alpha } \sqrt{p(1-p)} }{\sqrt{n} }

The 85% confidence level z_{\alpha } = 1.44

\sqrt{n} = \frac{Z_{\alpha } \sqrt{p(1-p)} }{m.E}

\sqrt{n} = \frac{1.44X\sqrt{0.3(1-0.3} }{0.04}

now calculation , we get

√n=13.80

now squaring on both sides n = 190.44

large sample n = 190.44≅190

<u>Conclusion</u>:-

Hence The  large  sample would be required in order to estimate the fraction of people who black out at 6 or more Gs at the 85% confidence level with an error of at most 0.04 is n = 190.44

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3 years ago
The temperature is -3.5 degrees Fahrenheit at 7:00 am. During the next 4 hours, the temperature decreases by -15.5 degrees. What
Kruka [31]
-19 degrees Fahrenheit
7 0
4 years ago
A box of 15 cookies cost 8 dollars how much dies it cosy to buy 1 cookie
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15÷8=1.875
I would round up to $1.88 /cookie.
4 0
4 years ago
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