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cestrela7 [59]
3 years ago
10

A manufacturer wants to double the volume of a 3 in.×2 in.×6 in. 3 i n . × 2 i n . × 6 i n . box, while using as little extra ca

rdboard as possible. Which statement is true?
Mathematics
1 answer:
viktelen [127]3 years ago
6 0

Answer: 2 inch dimension will give smallest increase.


Step-by-step explanation:

Length = 3 in

width = 2 in

height = 6 in

Extra cardboard  means to find surface area

on doubling the length

length = 6 In

width = 2 In

Height = 6In

Surface area for the above dimensions = 2 [ 6x2+2x6+6x6] = 120 sq in

On doubling the width

length = 3 in

width = 4 in

Height = 6 inch

Surface area for the above dimensions= 2 [ 3x4+4x6+6x3] = 2[54] = 108 sq inches

On doubling height

Length =3 in

width = 2 in

Height = 12 in

Surface area for above dimensions = 2 [ 3x2+2x12+12x3] = 2[6+24+36] = 132 sq inch

On doubling width surface area is minimum.

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A consumer products company is formulating a new shampoo and is interested in foam height (in mm). Foam height is approximately
Genrish500 [490]

Answer:

a) 0.057

b) 0.5234

c) 0.4766

Step-by-step explanation:

a)

To find the p-value if the sample average is 185, we first compute the z-score associated to this value, we use the formula

z=\frac{\bar x-\mu}{\sigma/\sqrt N}

where

\bar x=mean\; of\;the \;sample

\mu=mean\; established\; in\; H_0

\sigma=standard \; deviation

N = size of the sample.

So,

z=\frac{185-175}{20/\sqrt {10}}=1.5811

\boxed {z=1.5811}

As the sample suggests that the real mean could be greater than the established in the null hypothesis, then we are interested in the area under the normal curve to the right of  1.5811 and this would be your p-value.

We compute the area of the normal curve for values to the right of  1.5811 either with a table or with a computer and find that this area is equal to 0.0569 = 0.057 rounded to 3 decimals.

So the p-value is  

\boxed {p=0.057}

b)

Since the z-score associated to an α value of 0.05 is 1.64 and the z-score of the alternative hypothesis is 1.5811 which is less than 1.64 (z critical), we cannot reject the null, so we are making a Type II error since 175 is not the true mean.

We can compute the probability of such an error following the next steps:

<u>Step 1 </u>

Compute \bar x_{critical}

1.64=z_{critical}=\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}

\frac{\bar x_{critical}-\mu_0}{\sigma/\sqrt{n}}=\frac{\bar x_{critical}-175}{6.3245}=1.64\Rightarrow \bar x_{critical}=185.3721

So <em>we would make a Type II error if our sample mean is less than 185.3721</em>.  

<u>Step 2</u>

Compute the probability that your sample mean is less than 185.3711  

P(\bar x < 185.3711)=P(z< \frac{185.3711-185}{6.3245})=P(z

So, <em>the probability of making a Type II error is 0.5234 = 52.34% </em>

c)

<em>The power of a hypothesis test is 1 minus the probability of a Type II error</em>. So, the power of the test is

1 - 0.5234 = 0.4766

3 0
3 years ago
Max was on vacation twice as long as Jerad and half as long as wesley the boys were on vacation for a total of 3 weeks of long o
netineya [11]
Jerrad has a 3-day vacation, Max has a 6-day vacation, and Wesley's vacation is 12 days
3 0
3 years ago
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uysha [10]

Answer:

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Step-by-step explanation:

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8 0
2 years ago
As an unenrolled tax preparer, Daryl is subject to:
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I think there are missing details in this question. I'll just discuss what an unenrolled tax preparer is.

Unenrolled Tax Preparers are individuals who <span>possess the minimum qualifications required to prepare federal taxes and are granted a Preparer Tax Identification Number (PTIN) but are not Certified Public Accountants (CPA), lawyers, or enrolled agents.

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3 0
3 years ago
Read 2 more answers
Simplify your answer to the previous part and enter a differential equation in terms of the dependent variable xx satisfied by x
rusak2 [61]

Answer:

x'-5x=0, or x''-25x=0, or x'''-125x=0

Step-by-step explanation:

The function x(t)=e^{5t} is infinitely differentiable, so it satisfies a infinite number of differential equations. The required answer depends on your previous part, so I will describe a general procedure to obtain the equations.

Using rules of differentiation, we obtain that x'(t)=5e^{5t}=5x \text{ then }x'-5x=0. Differentiate again to obtain, x''(t)=25e^{5t}=25x=5x' \text{ then }x''-25x=0=x''-5x'. Repeating this process, x'''(t)=125e^{5t}=125x=25x' \text{ then }x'''-125x=0=x'''-25x'.

This can repeated infinitely, so it is possible to obtain a differential equation of order n. The key is to differentiate the required number of times and write the equation in terms of x.

7 0
3 years ago
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