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3 years ago

Solve 2(b − 3) ≥ 3(3 − b) and describe the graph of the solution. Show all your work.

Mathematics

1 answer:

solong [7]3 years ago

4 0

Answer:

$\large\boxed{b\geq3\to b\in[3,\ \infty)}$

Step-by-step explanation:

$2(b-3)\geq3(3-b)\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\(2)(b)+(2)(-3)\geq(3)(3)+(3)(-b)\\\\2b-6\geq9-3b\qquad\text{add 6 to both sides}\\\\2b-6+6\geq9+6-3b\\\\2b\geq15-3b\qquad\text{add}\ 3b\ \text{to both sides}\\\\2b+3b\geq15-3b+3b\\\\5b\geq15\qquad\text{divide both sides by 5}\\\\\dfrac{5b}{5}\geq\dfrac{15}{5}\\\\b\geq3$

$,\ \geq-\text{line to the right}\\-\text{o}\text{pen circle}\\\leq,\ \geq-\text{closed circle}$

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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.

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2 years ago

Hannah can type 450 words in 6 minutes. How many words per minute can she type?

Black_prince [1.1K]

Answer:

Step-by-step explanation:

That means she type 450/6 minutes. Take 450 and divide it by 6.

450/6 = 75

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3 years ago

when coloring in a grid to show a fraction, what does 3/10 look like when the grid represents a whole

Katarina [22]

Answer:

3 of 10 grid squares making up the whole will be colored.

Step-by-step explanation:

1/10 is one of ten equal parts. 3/10 is three of those ten equal parts. Essentially, the fraction 3/10 means "three out of ten".

_____

If the whole is divided into more than 10 parts, 3/10 will be 3 out of every 10. So if there are 20 parts, for example (2 tens), the 3/10 will be 6 of those parts (3 of each of the two tens).

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3 years ago

The value of a collectible coin can be represented by the equation mc018-1.jpg, where x represents its age in years and y repres

Scrat [10]

The answer is $53

x - age in years
y - total value in dollars

If the equation is y = 2x + 15, then after 19 years, the value will be:
x = 19

y = 2x + 15
y = 2 * 19 + 15
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3 years ago

Read 2 more answers

Pls Help me. There is 20 points to anyone who answers this first.

Alexeev081 [22]

bro doesnt it say the correct answer right there??

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