Answer:
1.625 kilograms
Step-by-step explanation:
Since total weight of pears and apples is 6.5 kilograms and 3/4 of this weight is pears, the weight of the pears is
→ 6.5 x 3/4 = 4.875 kilograms
Since the weight of the pears is 4.875, we can subtract it from the total weight to find the weight of the apples
→ 6.5 - 4.875 = 1.625 kilograms
Answer:
Answer is, 26.4
Step-by-step explanation:
By subtracting 51.82 by 26.37, you're left with a total of 26.37, and by rounding that to the tenths place, you get 26.4
Hi there
Down payment is 5% of the amount listed
100-95=5%
So the amount of down payment is
95,278×0.05=4,763.9
Hope it helps
Answer:
1
The probability is 
2
The probability is 
Step-by-step explanation:
From the question we are told that
The population mean is 
The standard deviation is 
The sample size is 
Generally the standard error for the sample mean
is mathematically evaluated as

substituting values


Apply central limit theorem[CLT] we have that
![P(\= X < 33) = [z < \frac{33 - \mu }{\sigma_{\= x}} ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3C%2033%29%20%3D%20%20%5Bz%20%3C%20%20%5Cfrac%7B33%20-%20%20%5Cmu%20%7D%7B%5Csigma_%7B%5C%3D%20x%7D%7D%20%5D)
substituting values
![P(\= X < 33) = [z < \frac{33 - 28.29 }{4.48} ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3C%2033%29%20%3D%20%20%5Bz%20%3C%20%20%5Cfrac%7B33%20-%20%2028.29%20%7D%7B4.48%7D%20%5D)
![P(\= X < 33) = [z < 1.05 ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3C%2033%29%20%3D%20%20%5Bz%20%3C%20%201.05%20%5D)
From the z-table we have that

For the second question
Apply central limit theorem[CLT] we have that
![P(\= X > 30 ) = [z > \frac{30 - \mu }{\sigma_{\= x}} ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3E%2030%20%29%20%3D%20%20%5Bz%20%3E%20%20%5Cfrac%7B30%20-%20%20%5Cmu%20%7D%7B%5Csigma_%7B%5C%3D%20x%7D%7D%20%5D)
substituting values
![P(\= X < 33) = [z > \frac{30 - 28.29 }{4.48} ]](https://tex.z-dn.net/?f=P%28%5C%3D%20X%20%3C%2033%29%20%3D%20%20%5Bz%20%3E%20%20%5Cfrac%7B30%20-%20%2028.29%20%7D%7B4.48%7D%20%5D)
From the z-table we have that

Thus

