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scZoUnD [109]
3 years ago
9

Without graphing, classify the following system as independent, dependent, or inconsistent.

Mathematics
1 answer:
Galina-37 [17]3 years ago
3 0

Answer:

Step-by-step explanation:

There is a missing operator in the second equation.

If the equation is 4x - 8y = 4, then the system is dependent. The second equation is a non-zero multiple of the first.

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An interior automotive supplier places several electrical wires in a harness.Apull test measures the force required to pull spli
oksano4ka [1.4K]

Answer:

a) For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assumin Normal distribution

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical dsitribution

The result is on the figure attached.

b) For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

Step-by-step explanation:

For this case we have the following data:

28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4

The quantile-quantile or q-q plot is a graphical procedure in order to check the validity of a distributional assumption for a data set. We just need to calculate "the theoretically expected value for each data point based on the distribution in question".

If the values are asusted to the assumed distribution, we will see that "the points on the q-q plot will fall approximately on a straight line"

For this case our distribution assumed is normal.

Part a

For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assuming Normal distribution (0,1)

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical distribution

The result is on the figure attached.

Part b

For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

4 0
4 years ago
Is 1.875 and rational number
jonny [76]

Answer: 1.875 is a rational number because it can be represented as a ratio or a fraction.

6 0
3 years ago
SOMEONE PLZ HELP WILL MARK BRAINLIEST TY
Pani-rosa [81]
-10.22, -7.89, -5.23, +34.98, +51.02, +51.22 because the closer a number is to 0 the smaller it is.
6 0
3 years ago
Read 2 more answers
Researchers fed mice a specific amount of Toxaphene, a poisonous pesticide, and studied their nervous systems to find out why To
Mkey [24]

Answer:

a. The mean refractory period= 1.85 and the standard error = 0.06455

b.  90% confidence interval for the mean absolute refractory period for all mice when subjected to the same treatment = 1.6981, 2.0019

c. Yes, the data give good evidence to support this theory

Step-by-step explanation:

a.  The table below shows the calculations:

                 X                (X-mean)^2

                1.7             0.0225

                1.8                 0.0025

                1.9                 0.0025

                2.0                 0.0225

Total        7.4                  0.05

Sample size: n=4

The mean is:  \bar{x} = \frac{7.4}{4} = 1.85

The sample standard deviation, s = \sqrt{\frac{\sum \left ( x-\bar{x} \right )^{2}}{n-1}}=0.1291

The standard error, se= \frac{s}{\sqrt{n}}=\frac{0.1291}{2} = 0.06455

b. Degree of freedom: df = n-1 = 3

Critical value of t for 90% confidence interval is: 2.3534

The confidence interval is  \bar{x}\pm t_{c}se = 1.85\pm 2.3534\cdot 0.06455=1.85\pm 0.1519 = (1.6981, 2.0019)

c. The Hypotheses are:

H_{0}:\mu=1.3,H_{1}:\mu>1.3

So the test statistics will be

t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=8.52

The p-value is: 0.0017

We reject the null hypothesis because p-value is less than 0.05 . This indicates that the data gave good evidence to support this theory.

4 0
3 years ago
A positive integer squared plus 3 times it’s consecutive integer is equal to 43
Eduardwww [97]

Answer:

the integer is 5

Step-by-step explanation:

x^2+3(x+1)=43

x^2+3x+3=43

x^2+3x+(9/4)=43-3+(9/4)

(x+(3/2))^2=42 1/4

x+(3/2)=sqrt(42 1/4)

x+(3/2)=6 1/2

x=(6 1/2)-(3/2)

x=5

check:

5^2+3(6)=43

25+18=43

43=43

3 0
2 years ago
Read 2 more answers
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