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TEA
1 month ago
7

Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The ar

ticle "Shake, Rattle, and Squeeze: How Much Is Left in That Container?" (Consumer Reports, May 2009: 8) reported on an investigation of this issue for various con- sumer products. Suppose five 6.0 oz tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the fol- lowing data (consistent with what the cited article reported): .53, .65, .46, .50, .37. Does it appear that the true average amount left is less than 10% of the advertised net contents?
Mathematics
1 answer:
Inessa [12.5K]1 month ago
7 0

Answer:

Indeed. Ample evidence substantiates the assertion that the leftover toothpaste is considerably below 10% of the stated net amount.

To explain step-by-step:

The remaining toothpaste sample consists of: [.53,.65,.46,.50,.37].

This dataset has a size of n=5, with a mean of M=0.502 and a standard deviation of s=0.102.

M=\dfrac{1}{5}\sum_{i=1}^{5}(0.53+0.65+0.46+0.5+0.37)\\\\\\ M=\dfrac{2.51}{5}=0.502

s=\sqrt{\dfrac{1}{(n-1)}\sum_{i=1}^{5}(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.53-(0.502))^2+...+(0.37-(0.502))^2]}\\\\\\s=\sqrt{\dfrac{1}{4}\cdot [(0.001)+(0.022)+(0.002)+(0)+(0.017)]}\\\\\\ s=\sqrt{\dfrac{0.04188}{4}}=\sqrt{0.01047}\\\\\\s=0.102

The 10% of the advertised content calculates to:

0.10\cdot 6.0\;oz=0.6\:oz

Hypothesis testing regarding the population mean:

The assertion posits that the remaining toothpaste is substantially less than 10% of the advertised net amount.

Thus, the null hypothesis and alternative hypothesis can be stated as follows:

H_0: \mu=0.6\\\\H_a:\mu< 0.6

The significance threshold is set at 0.05.

To find the estimated standard error of the mean, we utilize the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.102}{\sqrt{5}}=0.046

Subsequently, we can compute the t-statistic as follows:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{0.502-0.6}{0.046}=\dfrac{-0.098}{0.046}=-2.148

The degrees of freedom applicable to this sample size is:

df=n-1=5-1=4

This is a left-tailed test with 4 degrees of freedom resulting in t=-2.148. Thus, the P-value is evaluated as (using a t-table):

P-value=P(t

Given that the P-value (0.049) falls below the significance threshold (0.05), we conclude that the effect is significant.

The null hypothesis gets rejected.

There is ample evidence that affirms the claim that the remaining toothpaste is considerably below 10% of the stated net content.

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Jordan prepares 200 name tags to use at a meeting. The number for each color of the name tag is described below.
Leona [12618]
The answer is A. 55. Explanation: For the 35%, take your 200 and divide it by 2 to yield two 100's. Since 35% equates to a portion of 100%, grab 35 from each 100 and sum them up to reach 70. For the proportion of 3/8, divide which results in 0.375 then multiply by 200 to ascertain that 75 are yellow tags. Adding 75 to 70 provides a total of 145, which when subtracted from 200 results in 55, indicating there are 55 red tags.
5 0
2 months ago
Maia had 2064 more beads than Jenny. After Maia used 144 beads to make a necklace, she had 5 times as many beads as Jenny. A) ho
tester [12383]

Answer:

A) Maia had 1920 beads more.

B) Maia had 2544 beads at first.

Step-by-step analysis:

Let x denote the beads with Jenny and y for Maia.

The information provided states that Maia possesses 2064 beads more than Jenny, which can be represented mathematically as:

y=x+2064...(1)

Additionally, after Maia made a necklace with 144 beads, she had five times more beads than Jenny.

This can also be formulated as an equation:

y-144=5x...(2)

A) Since Maia had 2064 beads more than Jenny before using 144 beads, we calculate her final bead count by subtracting 144 from 2064.

\text{Number of beads Maia had more than Jenny in the end}=2064-144

\text{Number of beads Maia had more than Jenny in the end}=1920

Thus, Maia is left with 1920 beads more than Jenny.

B) To solve this system of linear equations, we will utilize the substitution method.

By inserting equation (1) into equation (2), we arrive at:

x+2064-144=5x

x+1920=5x

x-x+1920=5x-x

1920=4x

Now, let's divide our equation by 4.

\frac{1920}{4}=\frac{4x}{4}

480=x

Now, we'll substitute x=480 into equation (1) to isolate y.

y=480+2064

y=2544

Conclusively, Maia initially had 2544 beads.

3 0
2 months ago
The probability that an individual is left-handed is 0.12. in a class of 39 students, what is the probability of finding five le
Svet_ta [12734]

We start with the following information:

p = probability = 0.12<span>
n = total number of students = 39 </span>

x = number of left-handers = 5<span>
u = mean = p * n = 4.68
σ = standard deviation = √(n*p*(1-p)) = √(39 * 0.12 * 0.88) = 2.03</span>

Finding the z score:

z = (x – u) / σ

<span> z = (5 – 4.68) / 2.03
</span>

z = 0.1576 = 0.16

<span>

</span>Applying standard tables for z gives the p value as:

p value = 0.5636 = 56.36%

 

Consequently, there is a 56.36% probability.

 

8 0
1 month ago
Read 2 more answers
. Andrew made an error in determining the polynomial equation of smallest degree whose roots are 3, 2+2i
PIT_PIT [12445]

Answer:

Error made by Andrew: He identified incorrect factors based on the roots.

Step-by-step explanation:

The roots of the polynomial consist of: 3, 2 + 2i, 2 - 2i. By the factor theorem, if a is a root of the polynomial P(x), then (x - a) must be a factor of P(x). According to this premise:

(x - 3), (x - (2 + 2i)), (x - (2 - 2i)) represent the factors of the polynomial.

<pBy simplification, we obtain:

(x - 3), (x - 2 - 2i), (x - 2 + 2i) as the respective factors.

This is where Andrew's mistake occurred. Factors should always be in the form (x - a), not (x + a). Andrew expressed the complex factors incorrectly, resulting in an erroneous conclusion.

Thus, the polynomial can be expressed as:

(x - 3)(x - 2 - 2i)(x - 2+2i)=0\\\\ (x-3)(x^{2}-2x+2xi-2x+4-4i-2xi+4i-4i^{2})=0\\\\ (x-3)(x^{2}-4x+4+4)=0\\\\ (x-3)(x^{2}-4x+8)=0\\\\ x^{3}-4x^{2}+8x-3x^{2}+12x-24=0\\\\ x^{3}-7x^{2}+20x-24=0

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3 months ago
Una cartulina de 20 pulgadas de ancho por 32 pulgadas de alto será utilizada para colocar un cartel. El margen en la parte super
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Response:

could I receive an English version

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