Bella bought 3.2 meters of ribbon to decorate her pencil case. each meter cost $0.46. how much did the ribbon cost?
2 answers:
we know that
Each meter of ribbon cost
so
by proportion
Find the cost of meters of ribbon
therefore
<u>the answer is</u>
The ribbon cost
Answer: The cost of the ribbon is $1.472.
Step-by-step explanation: Given that Bella bought 3.2 meters of ribbon to decorate her pencil case and each meter cost $0.46.
We are to find the cost of the ribbon.
We have
the cost of 1 meter of ribbon = $0.46.
So, the cost of 3.2 meters of the ribbon will be
Thus, the total cost of the ribbon is $1.472.
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Answer:
<em>H₀</em>: <em>μ</em>₁ = <em>μ</em>₂ vs, <em>Hₐ</em>: <em>μ</em>₁ > <em>μ</em>₂.
Step-by-step explanation:
A two-sample <em>z</em>-test can be performed to determine whether the claim made by the owner of pier 1 is correct or not.
It is provided that the weights of fish caught from pier 1 and pier 2 are normally distributed with equal population standard deviations.
The hypothesis to test whether the average weights of the fish in pier 1 is more than pier 2 is as follows:
<em>H₀</em>: The weights of fish in pier 1 is same as the weights of fish in pier 2, i.e. <em>μ</em>₁ = <em>μ</em>₂.
<em>Hₐ</em>: The weights of fish in pier 1 is greater than the weights of fish in pier 2, i.e. <em>μ</em>₁ > <em>μ</em>₂.
The significance level of the test is:
<em>α</em> = 0.05.
The test is defined as:
The decision rule for the test is:
If the <em>p</em>-value of the test is less than the significance level of 0.05 then the null hypothesis will be rejected and vice-versa.
Answer:
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears
Step-by-step explanation:
For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.
However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
So
What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears
This probability is the pvalue of Z when X = 82. So
has a pvalue of 0.0885.
8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears