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Inga [223]
3 years ago
13

Give the slope of each of the following lines Name a point on each line y+2=2/3 (x-4)

Mathematics
1 answer:
Effectus [21]3 years ago
4 0

Answer:

See answer below

Step-by-step explanation:

y + 2 = 2/3(x - 4)

y + 2 = 2/3x - 8/3

y = 2/3x - 8/3 - 2

y = 2/3x - 2 2/3 - 2

y = 2/3x - 4 2/3  or y = 2/3x - 14/3  

Slope = 2/3  

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3 0
3 years ago
Extreme cold and hot temperatures are known to affect the operation of electronic components. Winter is approaching and you are
Musya8 [376]

Answer:

A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Test statistic(Z) = 2.31

P-value = 0.0104

Step-by-step explanation:

Step 1

Null hypothesis: The average damaging temperature of nine iPods is 5°F

Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F

Step 2

Mean=5°F, Sd=3, df=n-1=9-1=8

The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306

Confidence Interval(CI) = (mean + or - t×sd/√n)

CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F

CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F

Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31

Step 3

Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896

P-value = 1 - 0.9896 = 0.0104

Step 4

Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

7 0
3 years ago
Jonah finished 24 math problems in
8_murik_8 [283]

Answer:

3 hours

Step-by-step explanation:

In an hour there are 60 minutes to find the unit rate of problems per minute you divide 60/24 and get 2.5 multiply 2.5 by 72 get 180 in minutes that is 3 hours

3 0
3 years ago
Read 2 more answers
The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true
Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478

<u>Part B</u>

P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767

<u>Part C</u>

\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971

6 0
2 years ago
82, 62, 95, 81, 89, 51, 72, 56, 97, 98, 79, 85 order from least to greatest find the The median of the lower half of the data an
Studentka2010 [4]

Answer:

The median of the lower half of the data was determined to be 67 while the median of the upper half of the data was determined to be 92.

Step-by-step explanation:

<u>Step 1:  Order from least to greatest</u>

51, 56, 62, 72, 79, 81, 82, 85, 89, 95, 97, 98

<u>Step 2:  Determine the median of the lower half of the data</u>

There are 12 numbers which means that there is going to be 6 numbers on both sides of the data.  So using the first 6 numbers we will be able to determine the median of the lower half of the data.

51, 56, 62, 72, 79, 81

Since there is an even amount of numbers, that means that we will need to find the middle between the 3rd and 4th number.  We can do that by adding them up together and dividing by 2.

(62 + 72) / 2 = 67

Step 3:  <u>Determine the median of the upper half of the data</u>

So using the first 6 numbers we will be able to determine the median of the upper half of the data.

82, 85, 89, 95, 97, 98

Since there is an even amount of numbers, that means that we will need to find the middle between the 3rd and 4th number.  We can do that by adding them up together and dividing by 2.

(89 + 95) / 2 = 92

Answer: The median of the lower half of the data was determined to be 67 while the median of the upper half of the data was determined to be 92.

5 0
2 years ago
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