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Mathematics

1 answer:

zhannawk [14.2K]3 years ago

3 0

Answer:

2.5 is 12PM temp and difference is 3.9

Step-by-step explanation:

-0.5+3 because it is warmer and this is 2.5 also both B and C can be done to get the correct answer. and that is 2.5--1.4 which is the same as 2.5+1.4 which is 3.9

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What 2/3 + 5/7 complete denominators

umka21 [38]

You have to find common denominators. We can look at 3 and 7 and we know that nothing X 3 would equal to seven. So, we have to multiply
2/3 X 7 + 5/7 X 3
When we do that, we get 14/21 + 15/21
So when we add, the answer is
29/21
When we look at that, we have to see if it can be reduced. We know that both 7 and 3 can go into 21 but can they go into 29? No, so therefore, the answer cannot be reduced
Final answer is 29/21

8 0

4 years ago

Read 2 more answers

The mean amount purchased by a typical customer at Churchill's Grocery Store is $27.50 with a standard deviation of $7.00. Assum

Schach [20]

Answer:

a) 0.0016 = 0.16% probability that the sample mean is at least $30.00.

b) 0.8794 = 87.94% probability that the sample mean is greater than $26.50 but less than $30.00

c) 90% of sample means will occur between $26.1 and $28.9.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .