The answer is d<span>iscrete.</span>
Answer:
$460
Step-by-step explanation:
let the number of melons be x
4x + 3x + 3x = 200 melons
10x = 200 so, x will be 20
4x = 80 melons
3x = 60
3x = 60
amount made by 1st child = 80 × $2 = 160$
the remaining melons are 60 + 60 = 120
amount collected: they sold them in groups of 2 and each group was $5. so,
120÷2 × $5 = $300
total amount = 300 + 160 = $460.
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:
5.93 x 10^4
Step-by-step explanation:
5.93 is the decimal of 5,930,000
there are 4 zeros so it leads to 10 to the 4th power
Answer: 1320
Step-by-step explanation:
Given,
HCF of two numbers = 40
Product of two numbers = 52800
To find,
LCM of two numbers =?
Formula required,
HCF × LCM = product of two numbers
Calculation,
Using formula
→ HCF × LCM = product of two numbers
→ ( 40 ) × LCM = 52800
→ LCM = 52800 / 40
→ LCM = 1320
Therefore,
LCM of two numbers would be 1320.