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2 years ago

10

The sound intensity of rustling leaves is 100 times the reference intensity. Use your graph to determine the sound intensity of

rustling leaves in decibels.

2 answers:

Olenka [21]2 years ago

8 0

Answer:

its 20

Step-by-step explanation:

wariber [46]2 years ago

4 0

The sound intensity of rustling leaves, y = 20 decibels

<u>Step-by-step explanation:</u>

y = 10logx

The sound intensity of rustling leaves is 100 times the reference intensity.

x = ratio of intensity of sound to reference to intensity

x = 100:1 or x = 100

y = 10 log(100) = 10 * 2 = 20 decibels

y = 20 decibels

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frozen [14]

Answer:

you will have 71 tulips in each garden and 60 roses in each or 131 flowers together

Step-by-step explanation:

8 0

3 years ago

The USDA conducted tests for salmonella in produce grown in California. In an independent sample of 252 cultures obtained from w

Marat540 [252]

Answer:

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

Step-by-step explanation:

From the question we are told that

The first sample size is $n_1 = 252$

The number that tested positive is $k_1 = 18$

The second sample size is $n_2 = 476$

The number that tested positive is $k_2 = 20$

The level of significance is $\alpha = 0.01$

Generally the first sample proportion is mathematically represented as

$\^ p _1 = \frac{k_1 }{ n_1 }$

=> $\^ p _1 = \frac{18 }{ 252 }$

=> $\^ p _1 = 0.071$

Generally the second sample proportion is mathematically represented as

$\^ p _2 = \frac{k_2 }{ n_2 }$

=> $\^ p _2 = \frac{20 }{ 476}$

=> $\^ p _2 = 0.042$

The null hypothesis is $H_o : p_1 - p_2 = 0$

The alternative hypothesis is $H_a : p_1 - p_2 \ne 0$

Generally the test statistics is mathematically represented

$z = \frac{ \^ p_1 - \^ p_2 - ( p_ 1 - p_2 )}{ \sqrt{\frac{\^ p_1 (1-\^ p_1)}{ n_1 } + \frac{\^ p_2 (1-\^ p_2)}{ n_2 } } }$

=> $z = \frac{ 0.071 - 0.042 - 0 }{ \sqrt{\frac{0.071 (1-0.071)}{ 252 } + \frac{0.042 (1-\^ 0.042)}{ 476 } } }$

=> $z = 1.56$

From the z table the area under the normal curve to the right corresponding to 1.56 is

$P(Z > 1.56 ) =0.05938$

Generally the p-value is mathematically represented as

$p-value = 2 * P(Z > 1.56 )$

=> $p-value = 2 * 0.05938$

=> $p-value = 0.1188$

From the value obtained we see that $p-value > \alpha$ hence

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There no sufficient evidence to show that the proportion of salmonella in the region’s water differs from the proportion of salmonella in the region’s wildlife

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2 years ago

one foot of water column produces a pressure of 0.433psi . a gauge at the bottom of a water tower reads 44 psi. how high is the

Klio2033 [76]

The answer to the psi is 19.6

7 0

2 years ago

Can someone check this for me thanks.

kirill115 [55]

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8 0

2 years ago

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kondor19780726 [428]

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5 0

3 years ago