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3 years ago

Is this the correct way

Mathematics

2 answers:

crimeas [40]3 years ago

5 0

Yes it is good i hope this helps you

Anon25 [30]3 years ago

3 0

yes it is hope this helps you

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Use the graph of the derivative of f to locate the critical points x0 at which f has neither a local maximum nor a local minimum

jok3333 [9.3K]

Critical points are where the derivative is 0, i.e. where it crosses the x - axis

The Critical points lies where the derivative is 0, while it crosses the x-axis, SO, in this case the choice 3 looks like best answer for this.

6 0

3 years ago

The only way to identify an intercept is as an ordered pair. True False

artcher [175]

False because there are other ways to identify an intercept

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3 years ago

How is 40 gummy bears for 5 dollars written as a fraction?

Marina86 [1]

40/5 because to find the answer you would divided them

8 0

3 years ago

Read 2 more answers

!!20 POINTS!!

Sidana [21]

Answer:33.415 through 33.585 Or A

Step-by-step explanation:

You simply take 33.5 and add .085 to 33.5 get your answer.then subtract .085 from 33.5

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

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

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

= [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