The √136 is between which two consecutive integers?

Mathematics

1 answer:

Lorico [155]3 years ago

3 0

Hey there!!

We have 11² = 121 and 12² = 144

√136 is a decimal number

√136 = 11.6

Hence, the number √136 would fall between 11 and 12 . And 11 and 12 are consecutive.

Hope my answer helps!

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James, a real estate appraiser is looking at homes in the 78681 zip code. He takes a random sample of 42 homes form the Austin M

vazorg [7]

Answer:

80% confidence interval for the proportion of homes that are "eclectic" is [0.59 , 0.78].

Step-by-step explanation:

We are given that James takes a random sample of 42 homes form the Austin MLS (that's where Realtors go for information) and finds that 29 of them are listed as "eclectic" style.

Firstly, the pivotal quantity for 80% 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 homes that are "eclectic" = $\frac{29}{42}$ = 0.69

n = sample of homes = 42

$\mu$ = population proportion of homes that are "eclectic"

Here for constructing 80% confidence interval we have used One-sample z proportion statistics.

So, 80% confidence interval for the population proportion, p is ;

P(-1.282 < N(0,1) < 1.282) = 0.80 {As the critical value of z at 10% level

of significance are -1.282 & 1.282}

P(-1.282 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.282) = 0.80

P( $-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.80

P( $\hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.80

80% confidence interval for p = [ $\hat p-1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.282 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-1.282 \times {\sqrt{\frac{0.69(1-0.69)}{42} } }$ , $0.69+1.282 \times {\sqrt{\frac{0.69(1-0.69)}{42} } }$ ]