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Reika [66]
2 years ago
12

how many numbers did bill wrote if he wrote all the natural numbers between w and v not including w and v

Mathematics
1 answer:
dexar [7]2 years ago
4 0

Answer:

v - w - 1

Step-by-step explanation:

Try an experiment with a small, specific set of numbers like:

4, 5, 6, 7, 8, 9, 10   w = 4 and v = 10.

There are 5 numbers in the list, not counting the first and last.  Is this related to the difference between v and w?  v - w = 6, one more than 5.

The number of natural numbers between  w  and  v, not including  w  and  v, is v - w - 1.

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60% of what equals 27? This is the question. "Of" always means multiply.

So... convert the percent to a decimal: .6x=27.

Divide: x=27/.6
x=45

Plug it back in to check: .6(45)=27
27=27.

Or: 27/45=.6
.6=.6

8 0
3 years ago
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(a)It takes 51
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a) Given :

51 pounds of seed  can be planted in = 9 acre field.

Per pound of seed = \frac{9 \ acre}{51 \ pounds \ of \ seed}

We can convert 9/51 in simplest fraction by dividing top and bottom by 3.

We get

\frac{9/3}{51/3}=\frac{3}{17}

<h3>Therefore, 3/17 of an acrs can be planted per pound of seed.</h3>

b) $10 amount = 16 pounds of sugar.

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<h3>Therefore, 1.6 pounds of sugar she got per dollar.</h3>
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3 years ago
The proportion of defective computers built by Byte Computer Corporation is 0.15. In an attempt to lower the defective rate, the
FinnZ [79.3K]

Answer:

a)<em> Null hypothesis : H₀</em>:  the proportion of defective item of computer has been lowered. That is P < 0.15

<u><em>Alternative hypothesis: H₁:</em></u> The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b)    Test statistic   Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }

c)     Calculate the value of the test statistic = 0.991

d) The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e) Null hypothesis accepted at 0.01 level of significance

f) we accepted null hypothesis.

  Hence t<em>he proportion of defective item of computer has been lowered. </em>

Step-by-step explanation:

<u>Step(i)</u>:-

<em>Given the sample size 'n' = 42</em>

Given random sample of 42 computers were tested revealing a total of 4 defective computers.

The defective computers 'x' = 4

<em>The sample proportion of defective computers </em>

                                                                p = \frac{x}{n} = \frac{4}{42} = 0.095

<em>Given The Population proportion 'P' = 0.15</em>

<em>The level of significance ∝=0.01</em>

<u>Step(ii)</u>:-

a)<em> Null hypothesis : H₀</em>:  the proportion of defective item of computer has been lowered. That is P < 0.15

<u><em>Alternative hypothesis: H₁:</em></u> The proportion of defective item of computer

has been higher. That is P> 0.15 (Right tailed test)

b)

    Test statistic   Z = \frac{p-P}{\sqrt{\frac{PQ}{n} } }

                       

c)      

                 Z = \frac{0.095-0.15}{\sqrt{\frac{0.15(0.85)}{42} } }

                 z = \frac{-0.055}{\sqrt{0.00303} } = - 0.9991      

                     

  Calculate the value of the test statistic Z = - 0.9991

                                   |Z| = |- 0.9991| = 0.991

<u>Step(iii)</u>:-

d)

        The critical value at 0.01 level of significance = Z₀.₀₁ = 2.57

e)   Calculate the value of the test statistic Z = 0.991 < 2.57  at 0.01 level of significance.

<u><em>Conclusion</em></u>:-

    Hence the null hypothesis is accepted at 0.01 level of significance.

f)

<em>     The proportion of defective item of computer has been lowered.</em>

 

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3 years ago
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Step-by-step work:

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