If log8^x = p express log2^x in terms of p

Simplify the following rational expression

madreJ [45]

Step-by-step explanation:

as the denominator (x+3) is in both you need only one denominator to be multiplied in first rational number .

2x (x-2). + 5.

(x+3) (x-2). (x+3) (x-2)

2x^2-4x +5

(x+3) (x-2)

it's cannot be solved after that.

8 0

3 years ago

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John has a drawer full of socks. In his drawer, he has 5 pair of black socks, 6 pair of blue socks, 3 pair of white socks and 1

sukhopar [10]

Answer:

0.4

Step-by-step explanation:

Probability = Favourable Outcomes / Total Outcomes

Probability (Blue socks drawn out) = Blue Socks pairs / Total Socks pairs

6 / ( 5 + 6 + 3 + 1 )

= 6 / 15

= 0.4

8 0

3 years ago

A market research company conducted a survey to find the level of affluence in a city. They defined the category "affluence" for

malfutka [58]

Answer:

A 95% confidence interval for this population proportion is [0.081, 0.159].

Step-by-step explanation:

We are given that a market research company conducted a survey to find the level of affluence in a city.

Out of 267 persons who replied to their survey, 32 are considered affluent.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who are considered affluent = $\frac{32}{267}$ = 0.12

n = sample of persons = 267

p = population proportion

Here for constructing a 95% confidence interval we have used One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.12-1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ , $0.12+1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ ]

= [0.081, 0.159]

Therefore, a 95% confidence interval for this population proportion is [0.081, 0.159].

4 0

3 years ago

I need it asap fast and I give brainliest to a right answer.

katovenus [111]

Answer:

hiik

Step-by-step explanation:

rvnlgsudffstdrestdftuisffjhigyhoffyxhcccdfidsudjifg

8 0

2 years ago

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3(2x−7)+ 1/3 (7x+2) Please help Quick

atroni [7]

Simplified answer 25x-61/3

5 0

3 years ago