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3 years ago

8

Jonah runs 3/5 miles on Sunday and 7/10 miles on Monday.He uses the model to find that he ran a total of 1 mile. What mistake do

es Jonah make

1 answer:

antoniya [11.8K]3 years ago

7 0

Answer: He did not convert the fractions into the like fraction to add.

Step-by-step explanation:

Given: The distance Jonah runs on Sunday= $\frac{3}{5}$ miles

The distance Jonah runs on Monday= $\frac{7}{10}$ miles

The total distance he ran = $\frac{3}{5}+\frac{7}{10}\ miles$

Since both the fractions are not like , thus multiply 2 to the numerator and the denominator of the first fraction [to add fractions first convert them into like fractions], we get

The total distance he ran = $\frac{3\times2}{5\times2}+\frac{7}{10}\ miles$

$=\frac{6}{10}+\frac{7}{10}=\frac{6+7}{10}\\\\=\frac{13}{10}=1\frac{3}{10}\ miles$

The right answer is "The total distance he ran = $1\frac{3}{10}\ miles$ "

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3 years ago

-8(1 + 5b) – 2(86 – 1) = 50

kati45 [8]

Answer:

b = -57/10

Step-by-step explanation:

Let's first distribute everything out:

-8(1 + 5b) = -8 * 1 + (-8) * 5b = -8 - 40b

2(86 - 1) = 2 * 85 = 170

So, we have:

-8 - 40b - 170 = 50

-40b - 178 = 50

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Thus, the answer is b = -57/10.

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4 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

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

<u>Part I</u> : 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

<u />

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

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

= [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].

<u>Part II</u> : 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

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

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

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

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

= [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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3 years ago

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Mama L [17]

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3 years ago

Read 2 more answers

suppose six of these people are men, and the other six are women. in how many ways can they join hands for a circle dance, assum

Drupady [299]

Twelve people join hands for a circle dance.In how many ways can they do this? Suppose six of these people are men, and the other six are women. In how many ways can they join hands for a circle dance, assuming they alternate in gender around the circle

Answer:

86400 ways

Step-by-step explanation:

Since the circle can be rotated, the number of ways to arrange a distinct number of n objects in a circle will be (n−1)!.

Now, if we rotate the circle with the six women, we will see that there are 5! ways with which they can be placed in the circle.

After picking the places for the women, we will now fill each gap between two women with a man.

We have 6 men. Thus, number of ways to arrange the men is 6!

Thus,number of ways they can join hands for a circle dance, assuming they alternate in gender around the circle = 5! × 6! = 86400 ways

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3 years ago