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Leno4ka [110]
3 years ago
6

Jonathan wants to save up enough money so that he can buy a new sports equipment set that includes a football,baseball,soccer ba

ll ,and basketball. this complete boxed set costs $50 jonathan has $15 he saved from his birthday. in order to make more money, he plans to wash neghbors windows. he plans to charge $3 for each window he washes and any extra money he makes beyond $50 he can use to buy the additional accessories that go with the box set. writ and solve an inequality that represents the number of windows jonathan can wash in order to save at least the minimum amount he needs to buy the boxed set
Mathematics
1 answer:
soldi70 [24.7K]3 years ago
5 0

Answer:

Step-by-step explanation:

x = number of windows being washed

3x + 15 > = 50 (thats greater then or equal to)

3x > = 50 - 15

3x > = 35

x > = 35/3

x > = 11 2/3

so he would have to wash 12 windows to get the minimum amount.....any more window money would go to the additional accessories.

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Yes, it simplifies to 1/6.

Step-by-step explanation:

Both the top (numerator) and bottom (denominator) can be both divided by 3.  3 divided by 3 is 1. 18 divided by 3 is 6.

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1 + 2 + (-5) + 4 <br><br> A . -6<br> B . -5 <br> C . 0<br> D . 2<br> E . 4<br> F . 12
Dovator [93]

Answer:

Option D. 2

Step-by-step explanation:

1+2+(-5)+4

= 1+2-5+4

= 1+2+4-5

= 7-5

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The answer is b.......
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Use the given degree of confidence and sample data to construct interval for the population proportion. Of 369 randomly selected
Alexxx [7]

Answer:

e)  (3.77%, 8.70%)

95% confidence interval for the percentage of all medical students who plan to work in a rural community

(3.83% , 8.69%)

95% confidence level for the population proportion who blame oil companies for the recent increase in gasoline prices

(0.406 , 0.454)

Step-by-step explanation:

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

Given random sample size 'n' = 369

Sample proportion

                              p = \frac{x}{n} = \frac{23}{369} = 0.0623

95% confidence intervals are determined by

(p^{-} - Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} })

(0.0623 - 1.96 \sqrt{\frac{0.0623(1-0.0623)}{369} } , 0.0623 + 1.96\sqrt{\frac{0.0623(1-0.0623)}{369} })

(0.0623 - 0.0246 , 0.0623 + 0.0246)

(0.0383 , 0.0869)

(3.83% , 8.69%)

95% confidence interval for the percentage of all medical students who plan to work in a rural community

(3.83% , 8.69%)

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

Given 43% of those polled blamed of companies the most for the recent increase in gasoline prices

sample proportion 'p' = 0.43

Given Margin of error (M.E) = 0.024

95% confidence intervals are determined by

(p^{-} - Z_{0.05} \sqrt{\frac{p(1-p)}{n} } , p^{-} + Z_{0.05} \sqrt{\frac{p(1-p)}{n} })

(0.43 - 0.024 } , 0.43 +0.024 )

(0.406 , 0.454)

<u>Final answer</u>:-

95% confidence level for the population proportion who blame oil companies for the recent increase in gasoline prices

(0.406 , 0.454)

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