Answer:
Step-by-step explanation:
<u>Let the capacity of tank is x</u>
- 1/5x + 9 = 7/10x
- 2/10x + 9 = 7/10x
- 5/10x = 9
- 1/2x = 9
- x = 18 gallons
Answer:
2y² + 9
---------------
15y³
Step-by-step explanation:
Start by identifying the LCD, and then change each fraction so that its denominator is the LCD.
Here the LCD is 15y³, which is evenly divisible by 15y and 5y³.
Focus now on the first fraction: 2 / (15y). Multiplying numerator and denominator of this fraction by y² results in
y²·2 2y²
--------- → ----------
y²·15y 15y³ ←This is the correct LCD
Multiplying numerator and denominator of the second fraction by 3 results in:
3·3 9
------------ → ---------
3·5y³ 15y³ ←This is the correct LCD
So now those two original terms look like:
2y² 9
--------- + --------
15y³ 15y³
and this can be written in simpler form as:
2y² + 9
---------------
15y³
Answer:
The value is 
The correct option is a
Step-by-step explanation:
From the question we are told that
The margin of error is E = 0.05
From the question we are told the confidence level is 95% , hence the level of significance is
=> 
Generally from the normal distribution table the critical value of is
Generally since the sample proportion is not given we will assume it to be
Generally the sample size is mathematically represented as
![n = [\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cfrac%7BZ_%7B%5Cfrac%7B%5Calpha%20%7D%7B2%7D%20%7D%7D%7BE%7D%20%5D%5E2%20%2A%20%5C%5E%20p%20%281%20-%20%5C%5E%20p%20%29%20)
=> ![n = [\frac{ 1.96 }{0.05} ]^2 *0.5 (1 - 0.5)](https://tex.z-dn.net/?f=n%20%3D%20%5B%5Cfrac%7B%201.96%20%7D%7B0.05%7D%20%5D%5E2%20%2A0.5%20%281%20-%200.5%29%20)
=>
Generally the margin of error is mathematically represented as
Generally if the level of confidence increases, the critical value of 
increase and from the equation for margin of error we see the the critical value varies directly with the margin of error , hence the margin of error will increase also
So If the confidence level is increased, then the sample size would need to increase because a higher level of confidence increases the margin of error.
Shifted 5 units to the left.
A rectangle, presumably, as the cross section is going to be whatever shape the outward faces are (in the case of a vertical cross section like this one)
