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Montano1993 [528]

2 years ago

9

Abel’s age in 2019 will be equal to the sum of the digits of his year of birth. If Abel is an adult, how old is Abel now?

1 answer:

klasskru [66]2 years ago

3 0

Answer:

23

Step-by-step explanation:

If Abel was born in 1994, in 2019 he would be 23.

1+9+9+4=23

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Galen wrote the statement "If the sum of the digits in a number is divisible by 3, the original number is divisible by 3." Which

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the answer is B. good luck

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Fill in the gaps in the arithmetic sequence -3, _ ,_ , _, _, _ 12

lord [1]

Answer:

-0.5, 2, 4.5, 7, 9.5

Step-by-step explanation:

The given terms are 6 apart, so the common difference is 1/6 of their difference:

d = (12 -(-3))/6 = 15/6 = 5/2 = 2.5

Add 2.5 to each term to get the next one. Then the sequence is ...

-3, <u>-0.5</u>, <u>2.0</u>, <u>4.5</u>, <u>7.0</u>, <u>9.5</u>, 12

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2 years ago

The U.S. Energy Information Administration (US EIA) reported that the average price for a gallon of regular gasoline is $2.94. T

Anit [1.1K]

Answer:

a) 25

b) 67

c) 97

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample. In this problem, $\sigma = 0.25$

(a) The desired margin of error is $0.10.

This is n when M = 0.1. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.1 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.1\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.1}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.1})^{2}$

$n = 24.01$

Rounding up to the nearest whole number, 25.

(b) The desired margin of error is $0.06.

This is n when M = 0.06. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.06 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.06\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.06}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.06})^{2}$

$n = 66.7$

Rounding up, 67

(c) The desired margin of error is $0.05.

This is n when M = 0.05. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.05 = 1.96*\frac{0.25}{\sqrt{n}}$

$0.05\sqrt{n} = 1.96*0.25$

$\sqrt{n} = \frac{19.6*0.25}{0.05}$

$(\sqrt{n})^{2} = (\frac{19.6*0.25}{0.05})^{2}$

$n = 96.04$

Rounding up, 97

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