The show has a large display board where visitors are encouraged to pin up their own instant camera pictures, taken during the e
vent. The board is shown below.
How many 6 cm by 6 cm pictures could fit on the wall?
How many 6 cm by 6 cm pictures could fit on the wall?
1 answer:
Answer:
<h2>1,800 pictures</h2>Step-by-step explanation:
Find the diagram attached below with its dimension.
The board is rectangular in nature with dimension of 3.6 m by 1.8 m wall.
Area of a rectangle = Length * Breadth
Area of the board = 3.6 m * 1.8 m
since 1m -= 100cm
Area of the board = 360cm * 180cm
Area of the board = 64,800cm²
If the dimension of a picture on the wall is 6cm * 6cm, the area of one picture fir on the wall = 6cm* 6cm = 36cm²
In order to know the amount of 6cm* 6cm pictures that will fit on the wall, we will divide the area of the board by the area of one picture as shown;
Number of 6cm by 6cm pictures that could fit on the wall
= 64, 800cm²/36cm²
= 1,800 pictures
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The question is incomplete. Here is the complete qeustion.
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.10 0.13 0.16 0.15 0.14 0.008 0.15
(a) Construct a 99% confidence interval for the mean nitrogen-oxide emissions of all cars.
(b) If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, based on the 99% confidence interval in (a), can we safely conclude that this requirement is being met?
Answer: (a) 0.089 ≤ μ ≤ 0.171
(b) No
Step-by-step explanation:
(a) To determine the confidence interval, first calculate the mean (X) and standard deviation (s) of the sample
X =
X = 0.13
s =
s = 0.029
The degrees of freedom is
N - 1 = 7 - 1 = 6
And since the confidence is of 99%:
α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005
The t-test statistics for is 3.707
(Value found in the t-distribution table)
Now, calculate Error:
E = .
E = 3.707.
E = 0.041
The interval will be:
0.13 - 0.041 ≤ μ ≤ 0.13+0.041
0.089 ≤ μ ≤ 0.171
(b) No, because according to the interval, the nitrode-oxide emissions range from 0.089 to 0.171, which is greater than required by EPA.
For this case, what we must do is calculate the speed of the runner.
We have then that the speed is given by:
Where,
d: distance
t: time
Substituting values we have:
Answer:
the runners unit rate is: