Answer:
See attachment below
Explanation:
Which system of linear inequalities is represented by the graph? y > x – 2 and y x + 1 y x + 1 y > x – 2 and y < x + 1
2 answers:
Answer:
The graph representing the linear inequalities is attached below.
Explanation:
The inequalities given are :
y>x-2 and y<x+1
For tables for values of x and y and get coordinates to plot for both equation.
In the first equation;
y>x-2
y=x-2
y-x = -2
The table will be :
x y
-2 -4
-1 -3
0 -2
1 -1
2 0
The coordinates to plot are : (-2,-4) , (-1,-3), (0,-2), (1,-1) ,(2,0)
Use a dotted line and shade the part right hand side of the line.
Do the same for the second inequality equation and plot then shade the part satisfying the inequality.
The graph attached shows results.
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Given:
Let the speed of sound be represented by 'v' then
v ∝ (1)
= 349 m/s
= 340 m/s
= 20°C = 273+20 = 293 K
Formulae used:
1) °C = K + 273
2) K = °C - 273
3) °F = 1.8°C + 32
4) °R = °F + 459.67
Solution:
From eqn (1),
=
= 278.08 K
Now, Usinf formula (1), (2), (3) and (4) respectively, we get
1) T = 293 K
2) T = 293 -278.8 = 5.08°C
3) T = 1.8(5.08) + 32=41.14°F
4) T = 41.14 + 459.67 = 500.81°R
Answer:
See explanation
Explanation:
Solution:-
- A study on compressive strength of a concrete was made. The distribution of compressive strength ( experimental testing ) was normally distributed with variance ( σ^2 ).
- A random sample of n = 12 specimens were taken and the mean compressive strength ( μ ) of 3500 psi was claimed.
- We are to test the claim made by the civil engineer regarding the mean compressive strength of the concrete. The data of compressive strength of each specimen from the sample is given below:
3273, 3229, 3256, 3272, 3201, 3247, 3267, 3237,
3286, 3210, 3265, 3273
- We will conduct the hypothesis whether the mean compressive strength of the concrete conforms to the claimed value.
Null hypothesis: μ = 3500 psi
Alternate hypothesis: μ ≠ 3500 psi
- The type of test performed on the sample data will depend on the application of Central Limit Theorem.
- The theorem states that the sample can be assumed to be normally distributed if drawn from a normally distributed population. ( We are given the population is normally distributed; hence, theorem applies )
- We will approximate the mean of the population ( μ ) with the sample mean ( x ), as per the implication specified by the theorem.
- The mean of the sample ( x ) is calculated as follows:
- Since, we are testing the average compressive strength of a concrete against a claimed value. Any value that deviates significantly from the claimed value is rejected. This corroborates the use of one sample two tailed test.
- The test value may be evaluated from either z or t distribution. The conditions for z-test are given below:
- The population variance is known OR sample size ( n ≥ 30 )
- The population variance is known; hence, we will use z-distribution to evaluate the testing value as follows:
- The rejection region for the hypothesis is defined by the significance level ( α = 0.01 ). The Z-critical value ( limiting value for the rejection region ) is determined:
Z-critical = Z_α/2 = Z_0.005
- Use the list of correlation of significance level ( α ) and critical values of Z to determine:
Z-critical = Z_0.005 = ± 2.576
- Compare the Z-test value against the rejection region defined by the Z-critical value.
Rejection region: Z > 2.576 or Z < -2.576
- The Z-test value lies in the rejection region:
Z-test < Z-critical
-27.24 < -2.576 .... Null hypothesis rejected
Conclusion: The claim made by the civil engineer has little or no statistical evidence as per the sample data available; hence, the average compressive strength is not 3500 psi.
- To construct a confidence interval for the mean compressive strength ( μ ) we need to determine the margin of error for the population.
- The margin of error (ME) is defined by the following formula:
Where,
- The ( Z* ) is the critical value for the defined confidence level ( CI ):
- The confidence interval and significance level are related and critical value Z* is as such:
α = 1 - CI , Z* = Z_α/2
- The critical values for ( CI = 99% & 95% ) are evaluated:
α = 1 - 0.99 = 0.01 , α = 1 - 0.95 = 0.05
Z* = Z_0.005 , Z* = Z_0.025
Z* = ± 2.58 , Z* = ± 1.96
- The formulation of Confidence interval is given by the following inequality:
[ x - ME < μ < x + ME ]
[ x - Z*√σ^2 / n < μ < x + Z*√σ^2 / n ]
- The CI of 95% yields:
[ 3251.33 - 1.96*√(1000 / 12) < μ < 3251.33 + 1.96*√(1000 / 12) ]
[ 3251.333 - 17.89227 < μ < 3251.33 + 17.89227 ]
[ 3233.44 < μ < 3269.23 ]
- The CI of 99% yields:
[ 3251.33 - 2.58*√(1000 / 12) < μ < 3251.33 + 2.58*√(1000 / 12) ]
[ 3251.333 - 23.552 < μ < 3251.33 + 23.552 ]
[ 3227.78 < μ < 3274.88 ]
- We see that the width of the confidence interval increases as the confidence level ( CI ) increases. This is due to the increase in critical value ( Z* ) associated with the significance level ( α ) increases.
Hi, you haven't provided the programing language in which you need the code, I'll explain how to do it using Python, and you can follow the same logic to make a program in the programing language that you need.
Answer:
#Python
array = [5,2,3,4,5,0,7,1,9,8,2,7]
for i in range(len(array)-1):
array[i] = array[i]+array[i+1]
print(array)
Explanation:
First, we create an array, then a for loop that iterates from the beginning of the array to the penultimate value, this avoids changes in the last value of the array, we modify the array elements, except the last one, by adding the current element and the next element, finally, we print the result.
