We can use partial products model to solve 6×17
here Multiplicand =6
Multiplier=17
First break Multiplier (17) into 1 ten and 7 ones as
17=10+7
Then Multiply 6 to both of the parts of 17 then add it,
6×17
=6(10+7)
=6×10+6×7 [Distributive property]
=60+42=102
Partial Product is a product obtained by multiplying the multiplicand by one digit of the multiplier when the multiplier has more than one digit.
Answer:
The mean is 15.93 ounces and the standard deviation is 0.29 ounces.
Step-by-step explanation:
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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.
7% of the bottles containing this soft drink there are less than 15.5 ounces
This means that when X = 15.5, Z has a pvalue of 0.07. So when X = 15.5, Z = -1.475.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-1.475 = \frac{15.5 - \mu}{\sigma}](https://tex.z-dn.net/?f=-1.475%20%3D%20%5Cfrac%7B15.5%20-%20%5Cmu%7D%7B%5Csigma%7D)
![15.5 - \mu = -1.475\sigma](https://tex.z-dn.net/?f=15.5%20-%20%5Cmu%20%3D%20-1.475%5Csigma)
![\mu = 15.5 + 1.475\sigma](https://tex.z-dn.net/?f=%5Cmu%20%3D%2015.5%20%2B%201.475%5Csigma)
10% of them there are more than 16.3 ounces.
This means that when X = 16.3, Z has a pvalue of 1-0.1 = 0.9. So when X = 16.3, Z = 1.28.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.28 = \frac{16.3 - \mu}{\sigma}](https://tex.z-dn.net/?f=1.28%20%3D%20%5Cfrac%7B16.3%20-%20%5Cmu%7D%7B%5Csigma%7D)
![16.3 - \mu = 1.28\sigma](https://tex.z-dn.net/?f=16.3%20-%20%5Cmu%20%3D%201.28%5Csigma)
![\mu = 16.3 – 1.28\sigma](https://tex.z-dn.net/?f=%5Cmu%20%3D%2016.3%20%E2%80%93%201.28%5Csigma)
From above
![\mu = 15.5 + 1.475\sigma](https://tex.z-dn.net/?f=%5Cmu%20%3D%2015.5%20%2B%201.475%5Csigma)
So
![15.5 + 1.475\sigma = 16.3 – 1.28\sigma](https://tex.z-dn.net/?f=15.5%20%2B%201.475%5Csigma%20%3D%2016.3%20%E2%80%93%201.28%5Csigma)
![2.755\sigma = 0.8](https://tex.z-dn.net/?f=2.755%5Csigma%20%3D%200.8)
![\sigma = \frac{0.8}{2.755}](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cfrac%7B0.8%7D%7B2.755%7D)
![\sigma = 0.29](https://tex.z-dn.net/?f=%5Csigma%20%3D%200.29)
The mean is
![\mu = 15.5 + 1.475\sigma = 15.5 + 1.475*0.29 = 15.93](https://tex.z-dn.net/?f=%5Cmu%20%3D%2015.5%20%2B%201.475%5Csigma%20%3D%2015.5%20%2B%201.475%2A0.29%20%3D%2015.93)
The mean is 15.93 ounces and the standard deviation is 0.29 ounces.
Answer:
The answer is D: 30 g.
Step-by-step explanation: I googled the answer to this question and it said that NHS reccomends that adults take around 30 g of fibre as a daily dose. If you take less, you may not get enough, but if you take more, you would possibly overdose. So, the answer is D: 30g.
It would be 4 business owners will gift their employees