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3 years ago

How much does cell phone weigh

Mathematics

2 answers:

ZanzabumX [31]3 years ago

8 0

well it depends what kind of phone you have

noname [10]3 years ago

5 0

In 2005 he average weight is 4 ounces, they have gotten bigger since, clearly, but that is the average weight.
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Which represents the solution set to the inequality 5.1(3 + 2.2x) > –14.25 – 6(1.7x + 4)?

Tanya [424]

Answer:

1. Simplify the inequality 5.1(3 + 2.2x) > -14.25 - 6(1.7x + 4):

15.3+11.22x>-14.25-10.2x-24 10;15.3+11.22x>-10.2x-38.75.

2. Separate terms with x and without x in different sides:

11.22x+10.2x>-38.75-15.3.

3. Add similar terms:

21.42x>-54.05.

4. Divide by 21.42:

x>-54.05/21.42

x>-2.5

Step-by-step explanation:

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3 years ago

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marshall27 [118]

The point ends at the xy-axis

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3 years ago

Express 11.6311.6311, point, 63 as a mixed number

VladimirAG [237]

I thing it is six hundred

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3 years ago

Write -0.5 repeating as a fraction in simplest form

mario62 [17]

The answer is 5/9.
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3 years ago

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Suppose a simple random sample of size 50 is selected from a population with σ=10σ=10. Find the value of the standard error of t

bogdanovich [222]

Answer:

a) $\sigma_{\bar x} = 1.414$

b) $\sigma_{\bar x} = 1.414$

c) $\sigma_{\bar x} = 1.414$

d) $\sigma _{\bar x} = 1.343$

Step-by-step explanation:

Given that:

The random sample is of size 50 i.e the population standard deviation =10

Size of the sample n = 50

a) The population size is infinite;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

b) When the population size N= 50000

n/N = 50/50000 = 0.001 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

c) When the population size N= 5000

n/N = 50/5000 = 0.01 < 0.05

Thus ; the finite population of the standard error is not applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma_{\bar x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma_{\bar x} = \dfrac{10}{\sqrt{50}}$

$\sigma_{\bar x} = 1.414$

d) When the population size N= 500

n/N = 50/500 = 0.1 > 0.05

So; the finite population of the standard error is applicable in this scenario;

Therefore;

The standard error is determined as:

$\sigma _{\bar x} = \sqrt{\dfrac{N-n}{N-1} }\dfrac{\sigma}{\sqrt{n} } }$

$\sigma _{\bar x} = \sqrt{\dfrac{500-50}{500-1} }\dfrac{10}{\sqrt{50} } }$

$\sigma _{\bar x} = 1.343$

7 0

3 years ago