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myrzilka [38]
3 years ago
11

The maximum acceptable level of a certain toxic chemical in vegetables has been set at 0.4 parts per million (ppm). A consumer h

ealth group measured the level of the chemical in a random sample of tomatoes obtained from one producer to determine whether the mean level of the chemical in these tomatoes exceeds the recommended limit. The hypotheses are H0: μ ≤ 0.4 ppm Ha: μ > 0.4 ppm where μ is the mean level of the chemical in tomatoes from this producer. Explain the meaning of a Type I error. A Type I error would occur if, in fact, μ > 0.4 ppm, but the results of the sampling fail to lead to that conclusion. A Type I error would occur if, in fact, μ ≤ 0.4 ppm, but the results of the sampling lead to the conclusion that μ > 0.4 ppm A Type I error would occur if, in fact, μ ≤ 0.4 ppm, but the results of the sampling fail to lead to rejection of that fact. A Type I error would occur if, in fact, μ > 0.4 ppm, and the results of the sampling lead to rejection of the null hypothesis that μ ≤ 0.4 ppm.
Mathematics
1 answer:
Elodia [21]3 years ago
7 0

Answer:

Null hypothesis : \mu \leq 0.4

Alternative hypothesis: \mu >0.4

And for this case a type of error I for this case would be reject the null hypothesis that the population mean is lower or equal than 0.4 when actually is true.

A Type I error would occur if, in fact, μ ≤ 0.4 ppm, but the results of the sampling lead to the conclusion that μ > 0.4 ppm

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Type I error, also known as a “false positive” is the error of rejecting a null  hypothesis when it is actually true. Can be interpreted as the error of no reject an  alternative hypothesis when the results can be  attributed not to the reality.  

Type II error, also known as a "false negative" is the error of not rejecting a null  hypothesis when the alternative hypothesis is the true. Can be interpreted as the error of failing to accept an alternative hypothesis when we don't have enough statistical power

Solution to the problem

For this case we are trying to check the following hypothesis:

Null hypothesis : \mu \leq 0.4

Alternative hypothesis: \mu >0.4

And for this case a type of error I for this case would be reject the null hypothesis that the population mean is lower or equal than 0.4 when actually is true.

A Type I error would occur if, in fact, μ ≤ 0.4 ppm, but the results of the sampling lead to the conclusion that μ > 0.4 ppm

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SVETLANKA909090 [29]

Answer:

6.51 cm

Step-by-step explanation:

Since the sphere causes the water level in the cylindrical container to rise and thus increase by its own volume, the volume of the sphere is V = 4πr³/3 where r = radius of sphere. The volume rise of the container is thus    V' = πR²h where R = radius of base of cylinder = 7 cm and h = height of water level = 7.5 cm.

Since V = V',

4πr³/3 = πR²h

dividing through by π, we have

4r³/3 = R²h

multiplying both sides by 3/4, we have

r³ = 3R²h/4

taking cube-root of both sides, we have

r = ∛(3R²h/4)

Substituting the values of the variables into the equation, we have

r = ∛(3(7 cm)² × 7.5 cm/4)

r = ∛(3 × 49 cm² × 7.5 cm/4)

r = ∛(1102.5cm³/4)

r = ∛(275.625 cm³)

r = 6.508 cm

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3 years ago
A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.
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Answer:

a) \frac{ds}{dt}= v(t) = 3t^2 -18t +15

b) v(t=3) = 3(3)^2 -18(3) +15=-12

c) t =1s, t=5s

d)  [0,1) \cup (5,\infty)

e) D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

f) a(t) = \frac{dv}{dt}= 6t -18

g) The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

Step-by-step explanation:

For this case we have the following function given:

f(t) = s = t^3 -9t^2 +15 t

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

\frac{ds}{dt}= v(t) = 3t^2 -18t +15

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

v(t=3) = 3(3)^2 -18(3) +15=-12

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

3t^2 -18 t +15 =0

We can divide both sides of the equation by 3 and we got:

t^2 -6t +5=0

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

(t-5)*(t-1) =0

And for this case we got t =1s, t=5s

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

t^2 -6t +5 >0

(t-5) *(t-1)>0

So then the intervals positive are [0,1) \cup (5,\infty)

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6

And if we replace we got:

D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

a(t) = \frac{dv}{dt}= 6t -18

Part g and h

The particle is speeding up (1,3) \cup (5,\infty)

And would be slowing down from [0,1) \cup (3,5)

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Answer:

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Step-by-step explanation:

1. 15.8 * 3 = 47.4 (This is the length)

2. Find half of the rectangle

47.4 + 15.8 (width) = 63.2

3. Find the whole perimeter

63.2 * 2 = 126.4

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