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Tema [17]
3 years ago
11

Which problem solving strategy would be best to solve this problem?To study for the spelling competition, Jose memorized 2 words

from the list. The next day, Jose memorized 4 words from the list. The third day, Jose memorized 8 words from the list. If Jose continues, how many words will he memorize on the tenth day A.look for a patternB.work backwardsC.write an equationD.guess, check a
Mathematics
1 answer:
KengaRu [80]3 years ago
5 0
C. Because its always good to have notes
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11

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4-1=3/11-7=4

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Hey... Ans is in the pic..

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A manufacturing company wants to use a sample to estimate the proportion of defective tennis balls that are produced by a machin
Kitty [74]

Answer:

The least number of tennis balls needed for the sample is 1849.

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population proportion is:

 CI=\hat p\pm z_{\alpha/2}\ \sqrt{\frac{\hat p(1-\hat p)}{n}}

The margin of error for this interval is:

 MOE= z_{\alpha/2}\ \sqrt{\frac{\hat p(1-\hat p)}{n}}

Assume that the proportion of all defective tennis balls is <em>p</em> = 0.50.

The information provided is:

MOE = 0.03  

Confidence level = 99%

<em>α</em> = 1%

Compute the critical value of <em>z</em> for <em>α</em> = 1% as follows:

 z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.58

*Use a <em>z</em>-table.

Compute the sample size required as follows:

 MOE= z_{\alpha/2}\ \sqrt{\frac{\hat p(1-\hat p)}{n}}

       n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)} }{MOE}]^{2}

          =[\frac{2.58\times \sqrt{0.50(1-0.50)} }{0.03}]^{2}\\\\=1849    

Thus, the least number of tennis balls needed for the sample is 1849.

8 0
3 years ago
I'LL MARK BRAINLIEST PLEASE HELP
Fudgin [204]
<h3>Answer:</h3>

The number is 72

<h3 /><h3>Step-by-step solution:</h3>

Its hard to explain this really. Its kind of like a trial-and-error. Multiplying numbers by 8 with the second digit being 5 more than the units digit.

It takes time but eventually, youll find the answer.

Plus, You can search it up since theres already a pre-existing answer to this same question.

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2 years ago
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