All categories

2 years ago

I WILL MARK BRAINIEST

Mathematics

1 answer:

kozerog [31]2 years ago

8 0

Answer:parallel

Step-by-step explanation:

You might be interested in

8579 = 20.5% of x<br> What is x

frutty [35]

Hope attachment helps (✿^‿^)

6 0

3 years ago

231 ( 40 + 15 ) = 231 x 40 + ------------ x -----------<br><br> which property is this

a_sh-v [17]

Answer:

12705

Step-by-step explanation:

The problem could be solved in two different methods:

231×(55)=12705

231×40+231×15=9240+3465=12705

6 0

2 years ago

How do you find the slope of (2, -6) & (-3, -4)

marta [7]

(-4)-(-6)/(-3)-2=

2/-5

4 0

3 years ago

Read 2 more answers

Integers<br>53 + (-26 )

-BARSIC- [3]

<h2>✨ Question ✨</h2>

53 + (-26 )

<h2>✨ Answer ✨</h2>

$\: \: \: \: \: \: \: \: \: \purple{ \boxed{ \boxed{ \huge{ \tt{ \: 37 \: }}}}}$

6 0

3 years ago

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population 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 = 0.345

n = sample of participants = 3532

p = population proportion

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

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

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

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]

= [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

6 0

3 years ago