Which graph showed exponential growth

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

3 years ago

5х + 34 = -2 (1 - 7x) help me solve it step by step !

nataly862011 [7]

5x+34=-2(1-7x)

Steps:

5x+34=-2+14x
5x+34-14x= -2
-9x + 34= -2
-9x = -2-34
-9x = -36
X= -36/-9

Answer: x= 4

6 0

3 years ago

Please please please help. thank you.

djverab [1.8K]

Lets get started :)

The First question:
Diameter = 20 ft
Radius = $\frac{1}{2}$ of 20 (diameter) = 20 ft

Area formula of a circle is
A = $\pi$ r²
= $\pi$ (10)²
=100 $\pi$ ft²
≈ 314 ft²

The answer will be the second option

The Second question:
radius of circle = 12mm divided into 20 sectors area

A = $\pi$ r²
= $\pi$ (12)²
= 144 $\pi$ ft²

Divide into 20 equal sector areas = $\frac{144 \pi }{20} = 7.2 \pi$

≈ 22.6 mm²

Your answer will be the third option

The Third Question:

90°, sector area = 36 $\pi$ , Radius = ?

$\frac{angle}{360} = \frac{sector}{ \pi (r)^2}$
$\frac{90}{360} = \frac{36 \pi }{ \pi r^2}$
$\pi$ and $\pi$ cancels out

We can now cross multiply
360 × 36 = 90r²
12960 = 90r²

Divide by 90 on either side

$\frac{12960}{90} = \frac{90r^2}{90}$
144 = r²

Take squareroot
$\sqrt{144} = x$
x = 12 in

Your answer will be the third option

4 0

3 years ago

Simplify(3 square root 3 x square root 5 square root of 3)Square root of negative two

Assoli18 [71]

Answer:

9√10 (nine square root ten)

Step-by-step explanation:

(3√3 .√5.√3.)√2

= (3√3*3.√5)√2

=(3*3.√5)√2

=(9√5) √2

=9√5*2

=9√10

8 0

3 years ago

Which of the following is a description of a greatest common factor.

Simora [160]

Don’t open the link it’s a hacking business Good luck.

3 0

3 years ago