Jesse needs to wrap the present shown below. How much wrapping paper would he need?

Mathematics

1 answer:

maxonik [38]2 years ago

4 0

6in would be the answer

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Find the perimeter and area of a rectangle with a length of 2 meters and a width of 3 meters

klasskru [66]

Let L and W be the length and width of the rectangle, respectively. To solve for the perimeter (P) and area (A), use the following formula:

P = 2 x (L + W)

A = L x W

Substituting the given value for the dimensions,

P = 2 x (2m + 3m) = 10 m

A = 2m x 3m = 6m^2

Therefore, the perimeter is 10m and the area is 6m^2.

4 0

2 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is: