Answer:
31 L
Step-by-step explanation:
Here is the complete question
An ambulance require diesel as fuel because of so many trips nearby pump purchase diesel in bulk for the ambulance there are three containers at the pump which contain 403 L 434 L and 465 L of diesel respectively the maximum capacity of measuring container that can be measured the diesel of the three containers an exact number of times is
Solution
To find the exact number of times in which we can measure the diesel in the three containers, since the capacity of each container is 403 L, 434 L and 465 L respectively, we find the prime factors of each number.
The prime factors of 403 = 13 × 31
The prime factors of 434 = 2 × 7 × 31
The prime factors of 465 = 3 × 5 × 31
The measuring capacity equals the highest common factor (H.C.F) of the three numbers. That is H.C.F of 403, 434, 465.
Since it can be seen here that the H.C.F is 31, then the maximum capacity of measuring container that can be measured the diesel of the three containers an exact number of times is 31 L
70.1 ft² of fabric is needed by Lacy to make the sail boat.
<h3>
Trigonometric ratio</h3>
Trigonometric ratio is used to show the relationship between the sides and angles of a right angled triangle.
Let b represent the base of the triangle and h the height, hence:
sin(60) = h / 18
h = 15.6 feet
sin(30) = b / 18
b = 9 feet
The area of triangle = (1/2) * b * h = 0.5 * 9 * 15.6 = 70.1 ft²
70.1 ft² of fabric is needed by Lacy to make the sail boat.
Find out more on Trigonometric ratio at: brainly.com/question/6459892
Answer:
what do you need
Step-by-step explanation:
Answer:
Step-by-step explanation:
Hello!
You have sample of n=7 with mean X[bar]= 1403 and standard deviation S=27 and are required to estimate the mean with a 95%CI.
Asuming this sample comes from a normal population I'll use a stuent t to estimate the interval (a sample of 7 units is too small for the standard normal to be accurate for the estimation):
[X[bar]±
*
]

[1403±2.365*
]
[1378.87;1427.13]
The margin of error is the semiamplitude of the interval and you can calculate it as:

With a confidence level of 95% you'd expect that the real value of the mean is contained by the interval [1378.87;1427.13], the best estimate of the mean value is expected to be ± 24.13 of 1403.
I hope it helps!