Answer
5 and 1360
Step-by-step explanation:
If you divide 20 by 4 it is 5
If you times 20 by 68 mins then it is 1360
Always divided the two numbers they give you to find the rate which is 5 and then you times 68 to 20 to get 1360
Hope this helps
Ok. 1/4 is 9/36 . you do this by seeing what will give you 36 when you have a 4 . and 9/9 when you divided both you get 0.25.
again we have 4/9 and we see what we can put in to get another 36 , but we have a 9, so it’s 4. 4/9 x 4/4 = 16/36 when you divide this you them both you get 0.44 so they’re equivalent.
so 0.25 is less than 0.44
basically it’s saying 25<44
so 1/4 is less than 4/9
Answer and Step-by-step explanation:(MAKE BRAINLIEST PLEASE!!)
Let the quantity of understudies going to the game be s and the quantity of non-understudies going to the game be n. The all out number of individuals going to the game is 500. Address the present circumstance in a condition as s + n = 500.
Every understudy pays $5 for confirmation. Along these lines, s understudies pay 5s dollars in confirmations. Each non-understudy pays $10 for affirmation. Along these lines, n non-understudies pay 10n dollars in affirmations. Thus, the all out affirmations gathered is 5s + 10n.
The aggregate sum gathered was $4,000. So the subsequent condition is 5s + 10n = 4,000.
The answer for x is 5. :))))
Answer:
The probability that the plane is oveloaded is P=0.9983.
The pilot should take out the baggage and send it in another plain or have less passengers in the plain to not overload.
Step-by-step explanation:
The aircraft will be overloaded if the mean weight of the passengers is greater than 163 lb.
If the plane is full, we have 41 men in the plane. This is our sample size.
The weights of men are normally distributed with a mean of 180.5 lb and a standard deviation of 38.2.
So the mean of the sample is 180.5 lb (equal to the population mean).
The standard deviation is:
![\sigma=\frac{\sigma}{\sqrt{N}} =\frac{38.2}{\sqrt{41}}=\frac{38.2}{6.4} =5.97](https://tex.z-dn.net/?f=%5Csigma%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7BN%7D%7D%20%3D%5Cfrac%7B38.2%7D%7B%5Csqrt%7B41%7D%7D%3D%5Cfrac%7B38.2%7D%7B6.4%7D%20%3D5.97)
Then, we can calculate the z value for x=163 lb.
![z=\frac{x-\mu}{\sigma}=\frac{163-180.5}{5.97}=\frac{-17.5}{5.97}= -2.93](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B163-180.5%7D%7B5.97%7D%3D%5Cfrac%7B-17.5%7D%7B5.97%7D%3D%20%20%20-2.93)
The probability that the mean weight of the men in the airplane is below 163 lb is P=0.0017
![P(\bar X](https://tex.z-dn.net/?f=P%28%5Cbar%20X%3C163%29%3DP%28z%3C-2.93%29%3D0.00169)
Then the probability that the plane is oveloaded is P=0.9983:
![P(overloaded)=1-P(X](https://tex.z-dn.net/?f=P%28overloaded%29%3D1-P%28X%3C163%29%3D1-0.0017%3D0.9983)
The pilot should take out the baggage or have less passengers in the plain to not overload.