All categories

3 years ago

How do you solve this problem?

Mathematics

2 answers:

Sholpan [36]3 years ago

5 0

Answer:

21. r; 2.83 d; 5.66

22. d; 13.4 c; 42.09

23. r; 5.3 c; 33.30

24. r; 0.49 d; 0.99

Step-by-step explanation:

Hope this helps ;)

Inessa05 [86]3 years ago

5 0

Answer:

21) r; 2.83 d; 5.66 22) d; 13.4 c; 42.09

23) r; 5.3 c; 33.30 24) r; 0.49 d; 0.99

Step-by-step explanation:

it says someone just answered while i was typing so hopefully this is right

You might be interested in

I keep struggling on this question can someone help me figure it out

cestrela7 [59]

Answer:

the last one

Step-by-step explanation:

3 0

3 years ago

Solve. 213−12x=−23 Enter your answer in the box.

julia-pushkina [17]

$213 -12x = -23\\\\\implies 12x = 213 +23 = 236\\\\\implies x = \dfrac{236}{12}= \dfrac{59}3 = 19.67$

4 0

3 years ago

A survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the resul

pav-90 [236]

Answer:

95% confidence interval for the proportion of the adults who were opposed to the death penalty is (0.668, 0.704).

Step-by-step explanation:

We are given that a survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the results below, where X refers to the number of the respondents who were in favor.

X = 1,790

N = 2,610

$\hat p$ = Sample proportion = X/N = 0.6858

Firstly, the pivotal quantity for 95% 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 = 0.6858

n = sample of respondents = 2,610

p = population proportion

Here for constructing 95% confidence interval we have used One-sample z proportion statistics.

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