A machine is deisgned to fill 16 ounce bottles of shampoo.

What mathematical strenght and skills did you develop in this lesson

ElenaW [278]

Answer:

I have learned a lot in math, although it was years ago on of my favorite things to learn was to multiply and divide. They were not only easy but also fun.

Step-by-step explanation:

6 0

3 years ago

What is 7/10 - 5/12???? PLEASE HELP

Anarel [89]

Answer:

17/60

Step-by-step explanation:

5 0

2 years ago

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elizabeth and her sister each go to the store to buy a piece of pizza for $1.25 and a soda for $0.85 each. How much do they spen

Radda [10]

11
125
85
------
210
2
---------
420
3
--------
$12.50

6 0

3 years ago

What is the quotient of 3227/555? What is the repetend?

slega [8]

Answer:

A repeating or recurring decimal is decimal representation of a number whose digits are periodic (repeating its values at regular intervals) and the infinitely repeated portion is not zero. It can be shown that a number is rational if and only if its decimal representation is repeating or terminating (i.e. all except finitely many digits are zero). For example, the decimal representation of

1

/

3

becomes periodic just after the decimal point, repeating the single digit "3" forever, i.e. 0.333.... A more complicated example is

3227

/

555

, whose decimal becomes periodic at the second digit following the decimal point and then repeats the sequence "144" forever, i.e. 5.8144144144.... At present, there is no single universally accepted notation or phrasing for repeating decimals.

The infinitely repeated digit sequence is called the repetend or reptend. If the repetend is a zero, this decimal representation is called a terminating decimal rather than a repeating decimal, since the zeros can be omitted and the decimal terminates before these zeros.[1] Every terminating decimal representation can be written as a decimal fraction, a fraction whose divisor is a power of 10 (e.g. 1.585 =

1585

/

1000

); it may also be written as a ratio of the form

k

/

2n5m

(e.g. 1.585 =

317

/

2352

). However, every number with a terminating decimal representation also trivially has a second, alternative representation as a repeating decimal whose repetend is the digit 9. This is obtained by decreasing the final (rightmost) non-zero digit by one and appending a repetend of 9. 1.000... = 0.999... and 1.585000... = 1.584999... are two examples of this. (This type of repeating decimal can be obtained by long division if one uses a modified form of the usual division algorithm.[2])

Step-by-step explanation:

4 0

3 years ago

Front housing for cell phones are manufactured in an injection molding process. The time the part is allowed to cool in the mold

astra-53 [7]

Answer:

Step-by-step explanation:

Hello!

The manufacturer suspects that the time a part for cell phones is allowed to cool in the mold influences the occurrence of a defect in the finished housing.

The claim is that the longer the part is left to cool down, the fewer defects appear.

To test this claim, two random samples of 20 parts each were taken, one was left to cool down 10sec and the other was left to cool down for 20sec. After the cooling time, each one was visually inspected and scored based on their appearance from 1 (lowest) to 10 (highest).

Resulting:

X₁: Score given to a manufactured part after a cooling time of 10 seconds.

n₁= 20 parts

X[bar]₁= 3.35

S₁= 2.01

X₂: Score given to a manufactured part after a cooling time of 20 seconds.

n₂= 20 parts

X[bar]₂= 6.50

S₂= 1.54

Assuming both variables have a normal distribution, and the population variances are equal, the statistic to use is a student t for two independent samples with pooled sample standard deviation.

If the longer the cooling time, the fewer defects appear in the part, then the mean score will be greater for the sample with higher cooling time.

The statistical hypotheses are:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

$t_{H_0}= \frac{((X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_1} } } ~~t_{n_1+n_2-2}$

$Sa^2= \frac{(n_1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}$

$Sa^2= \frac{19*4.04+19*2.37}{20+20-2} = 3.81$

Sa= 1.95

$t_{H_0}= \frac{(3.35-6.50)-0}{1.95\sqrt{\frac{1}{20} +\frac{1}{20} } } = -5.108$

This test is one tailed to the left and so is the p-value, you can reach the value as follow:

p-value: $P(t_{38}\leq -5.108)$ ≅ 0.00001

p-value < 0.00001

The p.value is less than the significance level, the decision is to reject the null hypothesis.

Using a level of significance of 5% there is enough statistical evidence to reject the null hypothesis, so you can say that the average inspection scores of the parts that cooled down for 10 seconds are less than the average inspections scores of the parts that cooled down for 20 seconds, which means that the longer the cooling time, the fewer defects appear in the cellphone housing.

*-*-*

95% CI for the difference of means μ₁ - μ₂

The formula for the confidence interval is:

[(X[bar]₁-X[bar]₂) ± $t_{n_1+n_2-2;1-\alpha /2} * (Sa \sqrt{\frac{1}{n_1} +\frac{1}{n_2} } )$ ]

$t_{n_1+n_2-2;1-\alpha /2} = t_{38;0.975}= 2.024$

[(3.35-6.50)±2.024*(1.95* $\sqrt{\frac{1}{20} +\frac{1}{20} }$ )]

[-4.40;-1.90]

With a confidence level of 95%, you'd expect that the true value of the difference between the inspection scores of the parts that cooled down for 10 seconds and inspections scores of the parts that cooled down for 20 seconds will be included in the interval [-4.40;-1.90].

I hope it helps!

3 0

3 years ago

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