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vovikov84 [41]
3 years ago
12

number from 1 to 99 is written on a card making a pack of 99 cards. What is the probability of randomly selecting a card where t

he number contains at least one digit 1
Mathematics
1 answer:
andriy [413]3 years ago
3 0

Answer:

(19/99) = 0.192

Step-by-step explanation:

Numbers 1 to 99, that is 99 numbers (obtained through the equation for the nth term of an AP)

L = a + (n-1) d

L = nth term = 99

a = first term = 1

n = number of terms = ?

d = common difference = 1

99 = 1 + (n-1)1

n = 99

Sample space = 99

The numbers that include at least, a 1 are

1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91

19 numbers.

Probability of randomly selecting a card where the number contains at least one digit 1 from 1 to 99 = (19/99) = 0.192

Hope this Helps!!!

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Let f(x)=x^2 and g(x)=(x-3)^2+7 match up the correct transormations that are needed to transform the graph of f(x) to the graph
pentagon [3]

The transformations are vertical translation 7 units up.horizontal translation 3 units to the left

We have given that the equations

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Vertically translating a graph is equivalent to shifting the base graph up or down in the direction of the y-axis. A graph is translated to k units vertically by moving each point on the graph k units vertically.

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5 0
1 year ago
Can anyone help with the statements and reasons have been on this for a while :/
Bumek [7]

Answer:

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4 0
3 years ago
a major metroplitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. they aske
Arisa [49]

Answer:

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 1600, \pi = 0.4

99% confidence level

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 - 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.3685

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4 + 2.575\sqrt{\frac{0.4*0.6}{1600}} = 0.4315

The 99% confidence interval for the proportion of readers who would like more coverage of local news is (0.3685, 0.4315).

7 0
3 years ago
The sum of 4 consecutive integers is 40 what is the third number in this sequence?
Dmitry [639]
Let's represent the numbers by : x, x+1, x+2, x+3
x+x+1+x+2+x+3=40                     The third number in the sequence is x+2
4x+6=40                                           8.5+2=10.5
4x+6-6=40-6                               Therefore the third number in the sequence is
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4x/4=34/4
x=8.5
3 0
3 years ago
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