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elena-14-01-66 [18.8K]
3 years ago
5

Need help with 6,7,8,9,10,11,12,15

Mathematics
2 answers:
UNO [17]3 years ago
8 0
6) 1.01X10^-5
7) 6.X10^-14
8) 550000.
9) 60700000.
10) .000204
11) .000004
12) 7X10^12 is larger
Ainat [17]3 years ago
7 0

Answer:

6. 1.01 x -10^5

7. 6 x -10^14

8. 550,000

9. 60,700,000

10. 0.000204

11. 0.0004

12. 7,000,000,000,000 > 3,500,000,000

     7,000,000,000,000 (or 7 x 10^12) is greater by 2,000 times

15. 10^3 and 10^4

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Solve for A R=x(A+B)
julia-pushkina [17]
R=x(A+B)
divide both sides by x
R/x=A+B
minus B both sides
(R/x)-B=A
A=(R/x)-B
5 0
3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
3 years ago
Helppp mee asap !!!!!!
weqwewe [10]
Hyfgjgfthkok Hendrix djdhfnf fjfjfnfj fnnnnf fjjfjfnf fjfjfnfj
3 0
3 years ago
Thad attempted 25 free throws and made 17 of them while Amy attempted 30 and made 20. Who shot the higher percentage?
ahrayia [7]

Answer: Thad-1.471

Amy-1.500

SO AMY SHOT WITH A HIGHER PERCENTAGE

Step-by-step explanation: brainliest would be appreciated

3 0
2 years ago
(WITHIN 2 HOURS PLEASEEE!!)
goldfiish [28.3K]

Answer:

an average of 1.3 inches

Step-by-step explanation:

Using the mean absolute deviation, it can be concluded that the daily rainfall volume differs from the mean by an average of 1.3 inches.

What is the mean absolute deviation of a data-set?

The mean of a data-set is given by the sum of all observations divided by the number of observations.

The mean absolute deviation of a data-set is the sum of the absolute value of the difference between each observation and the mean, divided by the number of observations.

The mean absolute deviation represents the average by which the values differ from the mean.

In this problem, the mean absolute deviation is of 1.3 inches, hence, it can be concluded that the daily rainfall volume differs from the mean by an average of 1.3 inches.

5 0
2 years ago
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