The solution to the algebraic expression is: 23e - 21g - 14j + 32
What are algebraic expressions?
Algebraic expressions are mathematical expressions that contain variables, coefficients, and arithmetic operations such as addition, subtraction, division, and multiplication.
Solving algebraic expressions are an important part of mathematics as it helps to improve the aptitude and solving skills of the students.
From the given information, we have;
23j - 21g + 20e - 13 + 52e - 37j + 45 - 49e
let's rearrange by taking the like terms to the same sides;
= 23j -37j - 21g + 20e + 52e - 49e - 13 + 45
= -14j - 21g + 23e + 32
= 23e - 21g - 14j + 32
Therefore, we can conclude that the solution to the algebraic expression is: 23e - 21g - 14j + 32
Learn more about solving algebraic expressions here:
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I believe the answer is false
Hope this helps
Answer:
The best estimate is greater than 1/2 but less than 3/4
Step-by-step explanation:
If you multiply 1/4 by 3/3 (which is also 1), you can change the denominator without changing the value. So 1/4 is equal to 3/12.
Since keith has 11/12 hours to play, and he has already played 3/12 hours, subtract 3/12 from 11/12 to get 8/12 hours. This is how much time he has left to play.
If you simplify 8/12 hours, you get 2/3 hours.
So the best estimate would be: greater than 1/2 but less than 3/4.
Answer:
(0,1) , (2,4) , (4,7)
Explanation:
convert the equation into slope-intercept form (this would be y=3/2x+1)
now plug in random numbers for x and solve for y by multiplying them by 3/2 and adding 1.
First we need to write the null and alternate hypothesis for this case.
Let x be the average number of text message sent. Then
Null hypothesis: x = 100
Alternate hypothesis: x > 100
The p value is 0.0853
If p value > significance level, then the null hypothesis is not rejected. If p value < significance level, then the null hypothesis is rejected.
If significance level is 10%(0.10), the p value will be less than 0.10 and we reject the null hypothesis and CAN conclude that:
The mean number of text messages sent yesterday was greater than 100.
If significance level is 5%(0.05), the p value will be greater than 0.05 and we cannot reject the null hypothesis and CANNOT conclude that:
The mean number of text messages sent yesterday was greater than 100.