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Harrizon [31]
3 years ago
8

How many numbers between 1 and 1000 do not contain the digits 8, and 9 ?

Mathematics
2 answers:
sammy [17]3 years ago
6 0

Answer:

998

Step-by-step explanation:

I haven't included the <em><u>1</u></em> or the <em><u>1000</u></em>

Liono4ka [1.6K]3 years ago
6 0

Answer:

There are 200 numbers between 1 and 1000  that do not contain the digits 8, and 9.

Step-by-step explanation:

1-100

=(8,9,18,19,28,29,38,39,48,49,58,59,68,69,78,79,88,89,98,99)=20

101-200

=(108,109,118,119,128,129,138,139,148,149,158,159,168,169,178,179,188,189,198,199)=

201-300=20

.......

100*10=1000

20*10=200

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Abena has 1.3kg of apples and plenty of the other ingredients.
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Answer:

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7 0
3 years ago
The conditional relative frequency table was generated using data that compares the favorite subjects of male and female student
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Answer: Using the information of the conditional relative frequency table, 123 students in the survey said that math was their favorite subject.


Solution:

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3 0
3 years ago
Read 2 more answers
It is thought that not as many Americans buy presents to celebrate Valentine's Day anymore. A random sample of 40 Americans yiel
evablogger [386]

Answer:

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

 (0.3672 , 0.7328)

Step-by-step explanation:

<u><em>Explanation:</em></u>-

Given Random sample size n =40

Sample proportion

                            p = \frac{x}{n} = \frac{22}{40} = 0.55

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

   

          (p-Z_{\alpha } \sqrt{\frac{pq}{n} } , p + Z_{\alpha } \sqrt{\frac{pq}{n} } )

         

The Z-value Z₀.₉₈ = Z₀.₀₂ = 2.326

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

    (0.55-2.326\sqrt{\frac{0.55 X0.45}{40} } , 0.55 + 2.326\sqrt{\frac{0.55 X0.45}{40} } )

  ( 0.55 - 0.1828 , 0.55 + 0.1828)

  (0.3672 , 0.7328)

         

6 0
3 years ago
Read 2 more answers
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