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3 years ago

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Mathematics

2 answers:

polet [3.4K]3 years ago

8 0

Answer:

The answer is 6 to 12

Step-by-step explanation:

There are 6 ducks, and 12 chickens so therefore the ratio of ducks to chickens is 6:12. Hope this helps! (Brainliest pleasee I need one more!!) <3

podryga [215]3 years ago

7 0

Answer:

Step-by-step explanation:

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A body builder consumed 4,400 calories yesterday. Today, he consumed 3,600 calories. What is the percent decrease in number of c

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Answer: The answer is D

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3 years ago

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The correct answer is number one, number four, and number five. Please park Brainliest if helpful. Thanks

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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.

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2 years ago

A researcher claims that the variation in the salaries of elementary school teachers is greater than the variation in the salari

guajiro [1.7K]

Answer:

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

$F_{critical} = 1.8608$

$F = 2.9085$

The decision rule is

Reject the null hypothesis

There is sufficient evidence to support the researchers claim

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 30$

The sample variance for elementary school is $s^2_1 = 8324$

The second sample size is $n_2 = 30$

The sample variance for the secondary school is $s^2_2 = 2862$

The significance level is $\alpha = 0.05$

The null hypothesis is $H_o : \sigma^2_1 = \sigma^2 _2$

The alternative hypothesis is $H_a : \sigma_1 ^2 > \sigma^2_2$

Generally from the F statistics table the critical value of $\alpha = 0.05$ at first and second degree of freedom $df_1 = n_1 - 1 = 30 - 1 = 29$ and $df_2 = n_2 - 1 = 30 - 1 = 29$ is

$F_{critical} = 1.8608$

Generally the test statistics is mathematically represented as

$F = \frac{s_1^2 }{s_2^2}$

=> $F = \frac{8324 }{2862}$

=> $F = 2.9085$

Generally from the value obtained we see that $F > F_{critical }$ Hence

The decision rule is

Reject the null hypothesis

The conclusion is

There is sufficient evidence to support the researchers claim

4 0

2 years ago

Can anyone help me please solving this maths problems not sure if. Thanks

Marrrta [24]

The reason you can only compare the numerators is because the denominators are the same- leaving little comparison to really make. The numerators will determine which fraction is greater or smaller.

Ex. 4/5 or 3/5

which is greater? Which is smaller?

It can be seen that the denominators are the same, now finally comparing the numerators will answer both questions.

please vote my answer brainliest. thanks!

5 0

3 years ago