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3 years ago

12

a construction company loads bricks weighing 12 pounds each on a pallet weighing 6 pounds which expression gives the weight of a

pallet containing n bricks?

1 answer:

iragen [17]3 years ago

4 0

6+12n=weight blah blah gotta have 20 characters

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Find limit as x approaches -3- of x/(sqrt(x^2-9)

gregori [183]

$lim_{x \rightarrow -3} \frac{x}{x^2-9} =lim_{x \rightarrow -3} \frac{1}{2x} = \frac{1}{2(-3)} =- \frac{1}{6}$

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3 years ago

How many $3.50 burgers can you buy for a million dollars? Give your answers to the nearest whole numbers.

Nookie1986 [14]

$\frac{1000000}{3.5}= 285714.29 = 285714$ Answer here Burgers

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3 years ago

Read 2 more answers

In an intro stats class, 57% percent of students eat breakfast in the morning and 80% of students floss their teeth. Forty-six p

WINSTONCH [101]

Answer: B. 11%

Step-by-step explanation:

Let A = Event that students eat breakfast in the morning.

B= Event that students floss their teeth.

We are given P(A)=57%=0.57

P(B)=80%=0.80

P(A∩B) = 46% =0.46

Now, the probability that a student from this class eats break feast but does not floss their teeth :-

$P(A-B)=P(A)-P(A\cap B)\\\\ = 0.57-0.46=0.11=11\%$

Hence, the probability that a student from this class eats break feast but does not floss their teeth= 11%

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3 years ago

A quality control expert wants to estimate the proportion of defective components that are being manufactured by his company. A

tigry1 [53]

Answer:

A sample of 1032 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A sample of 300 components showed that 20 were defective.

This means that $\pi = \frac{20}{300} = 0.0667$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

How large a sample is needed to estimate the true proportion of defective components to within 2.5 percentage points with 99% confidence?

A sample of n is needed.

n is found when M = 0.025. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.025 = 2.575\sqrt{\frac{0.0667*0.9333}{n}}$

$0.02\sqrt{n} = 2.575\sqrt{0.0667*0.9333}$

$\sqrt{n} = \frac{2.575\sqrt{0.0667*0.9333}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.0667*0.9333}}{0.02})^2$

$n = 1031.9$

Rounding up

A sample of 1032 is needed.

6 0

2 years ago

Find the product. If possible, rename the product as a mixed number or a whole number 7/2 x 9/10

tamaranim1 [39]

Step-by-step explanation:

7/9×9/10

63/20

3 3/20

.....

8 0

3 years ago