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3 years ago

How do you round 7,298,341 to the nearest 10,000.

1 answer:

3 0

You would have to underline 9 since that is the 10,000 place and draw an arrow from the 9 to the 8 next to it and if it is 5 or more you would have to add 1 to the 9 if it is 4 or less you would change the whole number into 7,290,000. If it is 5 or more the 9 would become a 10 which would make the number 7,300,000

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Peter and Henry went to the movies Peter bought the two Movie tickets for 7 dollars each. How much did he spend for the movie ti

ivann1987 [24]

Answer:

tickets: $14

popcorn: $10

total: $24

Step-by-step explanation:

2 movie tickets at $7 each is 2*7 which is $14

2 buckets of popcorn at $5 each is 2*5 which is $10

adding these together we get a total spending of $24

3 0

2 years ago

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

Can you please help me with the whole table

Bad White [126]

44,000
38,720
34,073.60
29,984.77
26,386.60

7 0

3 years ago

Find the sum of the first 8 terms of the following sequence. Round to the nearest hundredth if necessary

AlexFokin [52]

The sum of the first 8 terms is 2.51 to the nearest hundredth

Step-by-step explanation:

In the geometric sequence there is a constant ratio between each two consecutive terms

The formula of the sum of n terms of a geometric sequence is:

$S_{n}=\frac{a(1-r^{n})}{1-r}$ , where

a is the first term
r is the constant ratio between the consecutive terms

∵ The sequence is 6 , -5 , 25/6 , .............

∵ -5 ÷ 6 = $\frac{-5}{6}$

∵ $\frac{25}{6}$ ÷ -5 = $\frac{-5}{6}$

- There is a constant ratio between the consecutive terms

∴ The sequence is a geometric sequence

∵ The first term is 6

∴ a = 6

∵ The constant ratio is $\frac{-5}{6}$

∴ r = $\frac{-5}{6}$

∵ We need to find the sum of 8 terms

∴ n = 8

- Substitute the values of a, r and n in the rule above

∴ $S_{8}=\frac{6[1-(\frac{-5}{6})^{8}]}{1-(\frac{-5}{6})}$

∴ $S_{8}=2.511595508$

- Round it to the nearest hundredth

∴ $S_{8}=2.51$

The sum of the first 8 terms is 2.51 to the nearest hundredth

Learn more:

You can learn more about the sequences in brainly.com/question/7221312

#LearnwithBrainly

3 0

4 years ago

13 R S 12 What's the length of QR? A) 1 B) 17.7 C) 6.7 OD) 5

ANEK [815]

Answer:

Step-by-step explanation:

This is a right triangle, so we can use the Pythagorean theorem

a^2+b^2 = c^2

where a and b are the legs and c is the hypotenuse

QR^2 + 12^2 = 13^2

QR^2 +144 =169

QR^2 = 169-144

QR^2 =25

Take the square root of each side

QR = sqrt(25)

QR =5

8 0

3 years ago