Write your answer as a mixed number in simplest form

Mathematics

1 answer:

Neko [114]3 years ago

5 0

Answer:

2 5/12

Step-by-step explanation:

You put it so that the denominators are the same so in this case I used 12 as the denominator and used the rule for each. Next I subtracted 3 9/12 minus 1 4/12 and got 2 5/12

You might be interested in

một hộp đựng 11 tấm thẻ được đánh số từ 1 đến 11. tính số cách lấy ra 4 tấm thẻ từ hộp đó để tổng các số ghi trên 4 tấm thẻ ấy l

Whitepunk [10]

Answer:

16/33

Step-by-step explanation:

Tổng của cả 4 tấm thẻ là một số lẻ khi:

Có 1 thẻ là lẻ, 3 thẻ còn lại là chẵn, có: 6C1.5C3=60 cách

Có 3 thẻ là lẻ, 1 thẻ là chẵn, có: 5C1.6C3=100 cách

Suy ra có 160/11C4=16/33 cách

3 0

3 years ago

The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability

amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

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

Also, important to remember that the standard deviation is the square root of the variance.

Normal probability distribution:

Problems of normally distributed samples are solved using 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$