25m+100−24m−75=68
Step 1: Simplify both sides of the equation.
25m+100−24m−75=68
25m+100+−24m+−75=68
(25m+−24m)+(100+−75)=68(Combine Like Terms)
m+25=68
m+25=68
Step 2: Subtract 25 from both sides.
m+25−25=68−25
m=43
Answer:
m=43
Answer:
1 = No 2 = Yes 3 = 1.88
Step-by-step explanation:
Answer:
1 1 1 1
Step-by-step explanation:
Answer:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
Step-by-step explanation:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
Answer:
Bob forgot to distribute 3 to 1.50 in the parentheses
Step-by-step explanation:
Bob made the mistake in his first step.
In the equation 3(x + 1.50) = 30, the 3 has to be distributed to both x and 1.50.
But, Bob only distributed it to x, which led him to get the wrong answer.
So, Bob's mistake when solving was that he forgot to distribute the 3 to 1.50