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3 years ago

1+1= 2+2= 2 and 3 freee

2 answers:

8 0

<h2><u><em>Question- 1 + 1</em></u></h2><h2><u><em>Answer 2</em></u></h2><h2><u><em>Question 2+2 </em></u></h2><h2><u><em>Answer-4 </em></u></h2><h2><u><em>Question- 2+3</em></u></h2><h2><u><em>Answer- 5</em></u></h2>

solong [7]3 years ago

3 0

Answer:

2,4,5

Step-by-step explanation:

1+1=2 or l+l

2+2=4 or ll+ll

2+3=5 or ll+lll

thx for points

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Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500

Alborosie

Answer:

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

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:

$n = 500, \pi = \frac{421}{500} = 0.842$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 - 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.81$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 + 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.874$

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

6 0

3 years ago

Solve each equation (quadratic pattern)

dusya [7]

Answer: x = 2

<u>Step-by-step explanation:</u>

$2^{2x}-2^x-12=0\\\\\text{Let u = }2^x\\\\u^2-u-12=0\\(u-4)(u+3)=0\\\\u-4=0\quad and\quad u+3=0\\u=4\qquad and\quad u=-3\\\\\text{Substitute u with }2^x\\2^x=4\qquad and \quad 2^x=-3\\2^x=2^2\quad and\quad \text{not possible}\\\boxed{x=2}$

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Answer: x = 0

<u>Step-by-step explanation:</u>

$3^{2x}+3^{x+1}-4=0\\\\3^{2x}+3^x\cdot3^1-4=0\\\\\text{Let u = }3^x\\u^2+3u-4=0\\\\(u+4)(u-1)=0\\\\u+4=0\quad and\quad u-1=0\\u=-4\qquad and\quad u=1\\\\\text{Substitute u with }3^x\\3^x=-4\qquad and\quad 3^x=1\\\text{not possible}\ and\quad 3^x=3^0\\.\qquad \qquad \qquad \qquad \boxed{x=0}$

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Answer: No Solution

<u>Step-by-step explanation:</u>

$4^x+6\cdot 2^x+8=0\\\\2\cdot 2^x+6\cdot 2^x+8=0\\\\\text{Let u = }2^x\\2u+6u+8=0\\8u+8=0\\8u=-8\\u=-1\\\\\text{Substitute u with }2^x\\2^x=-1\\\text{not possible}$

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Answer: No Solution

<u>Step-by-step explanation:</u>

<u> $9^x=3^x-6\\\\3\cdot 3^x=1\cdot 3^x-6\\\\\text{Let u = }3^x\\\\3u=u-6\\2u=-6\\u=-3\\\\\text{Substitute u with }3^x\\3^x=-3\\\text{not possible}$ </u>