I need the answers for 1-6
1 answer:
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Answer:
12 black beads
Step-by-step explanation:
Given the ration of gold beads to black beads on the short necklace:
Since the ratio of 4:1 has 5 total parts, you can take the total amount of beads in the long necklace and divide by 5:
60 ÷ 5 = 12
12 beads represents 1 part of the ratio, so that would be the number of black beads. To check, multiply 12 by 4: 12 x 4 = 48, and set up a ratio:
48: 12 = 4: 1
The order of operations is defined by the acronym PEMDAS (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction).
Therefore
6 + [(-24) ÷ 6] / 19
= 6 + [ -24/6 ] / 19 handle parentheses first
= 6 + (-4) /19
= 6 - 4/19 handle division
= 110/19
= 5 5/9
Answer:
Therefore
6 + [(-24) ÷ 6] / 19
= 6 + [ -24/6 ] / 19 handle parentheses first
= 6 + (-4) /19
= 6 - 4/19 handle division
= 110/19
= 5 5/9
Answer:
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.