Solve for n. d=4m + 7n n
2 answers:
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Answer:
Alyssa had scored 3 goals today.
Step-by-step explanation:
Given:
Number of goals scored yesterday =
Total number of goals scored in 2 games = 4
We need to find the number of goals scored today.
Solution:
Now Given:
Today she scored 3 times as many goals than yesterday.
So we can say that;
Number of goals scored today =
Now we know that;
Total number of goals scored in 2 games is the sum of Number of goals scored yesterday and Number of goals scored today.
framing in equation form we get;
Dividing both side by 4 we get;
Number of goals scored yesterday = 1
Number of goals scored today =
Hence Alyssa had scored 3 goals today.
Answer:
0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are 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.
Central limit theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?
This is 1 subtracted by the pvalue of Z when X = 3.1. So
By the Central Limit Theorem
has a pvalue of 1.
1 - 1 = 0
0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.