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2 years ago

May somebody help me with this question? It is in the picture.

Mathematics

2 answers:

UNO [17]2 years ago

3 0

Answer:

Step-by-step explanation:

pychu [463]2 years ago

3 0

Answer: 40%

you can do a ration to help solve your roblem. 36/90 x/100 x=40
pastrami/total %/100

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There are 5,000 books in the town's library. Of these, 4,700 are fiction. To find the percent of the books that are fiction,

adell [148]

Answer:

94%

Step-by-step explanation:

4,700/5,000 x 100%

0.94 x 100% = 94%

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2 years ago

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Explain the difference between a whole number and an integer.

lesantik [10]

Answer:

Whole numbers are all natural numbers including 0 e.g. 0, 1, 2, 3, 4… Integers include all whole numbers and their negative counterpart e.g. … -4, -3, -2, -1, 0,1, 2, 3, 4,… Where a and b are both integers.

Step-by-step explanation:

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2 years ago

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Give all the basic trigonometric functions that satisfy the condition <img src="https://tex.z-dn.net/?f=f%28x%29%20%5Cneq%200" i

prohojiy [21]

Answer:

i need points lol

Step-by-step explanation:

sorryyyyy

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2 years ago

Penelope has an 8-ounced flavoured water. She drinks 2 ounces .

Nataliya [291]

She now has 6 oz. if that’s what you’re asking. If not please clarify

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2 years ago

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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:

Step(i):-

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%)

Step(ii):-

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)

Final answer:-

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

(0.406 , 0.454)

6 0

3 years ago