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2 years ago

Find the surface area of each figure. Round to the nearest tenth if necessary.

Mathematics

1 answer:

Andre45 [30]2 years ago

4 0

Answer:

365.7 km²

hopefully this answer can help you to answer the next question.

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Find the area of the trapezoid. b1=5 b2=7 h=4 PLEASE EXPLAIN TOO!

olga_2 [115]

Step-by-step explanation:

The formula for finding the area of a trapezoid is $\frac{b_{1} + b_{2} }{2}$ ×h

Start by substituting the values given in the problem into the formula

$\frac{5+7}{2}$ ×4

Now Simply/solve

$\frac{12}{2}$ ×4

6×4

The area of the trapezoid is 24 $units^{2}$

I hope this helps!!!

4 0

2 years ago

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Question 18 Item 18 The distribution of the number of moths captured per night by a certain moth trap is approximately normal wi

juin [17]

Answer:

Standard deviation 46.3

Step-by-step explanation:

A 28% of the distribution is equivalent to a z score of -0.5828, you can use a z table to find that.

Z score is calculated as follows:

$z = \frac{x-\mu}{\sigma}$

x is the number being evaluated

μ is the mean

σ is the standard deviation

And it is used to calculate how many standard deviations you are from the mean of the sample.

Replacing with the known information you can calculate the standard deviation:

$\sigma=\frac{x-\mu}{z} =\frac{76-103}{-0.5828}= 46.3$

4 0

3 years ago

The length of chord AB in D is 9mm. If the measure of angle AB is 32 then find the length of AB

Taya2010 [7]

The length of arc AB is 9.12 mm:

We first calculate for the radius r of the circle using the equation
r = c/(2 sin[theta/2])

where c is the length of chord AB that is given as 9 millimeters
angle given is 32 degrees

To convert theta 32 degrees into radians:
32 degrees * (pi/180) = 32 degrees * (3.14/180) = 0.5583 radians

We now substitute the values into the equation to find the radius r:
r = 9/(2 sin[0.5583/2])
r = 16.33 mm
.
We can finally solve for the length s of arc:
s = r theta = 16.33 * 0.5583 = 9.12 mm

5 0

3 years ago

46. Melanie wants to construct the perpendicular bisector of

vekshin1

Answer:

Step-by-step explanation:

7 0

2 years ago

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What are the coordinates of the endpoints of the midsegment for △JKL that is parallel to KL¯¯¯¯¯? Enter your answers in the boxe

labwork [276]

Answer:

J(-2, -1), K(4, -5), L(0, -5)

The midpoint of segment JL is

(-2 + 0)/2, (-1 + (-5))/2) = (-2/2, -6/2) = (-1, -3)

The midpoint is

(-2 + 4)/2, (-1 + (-5))/2) = (2/2, -6/2) = (1, -3)

A: (-1, -3), (1, -3)

Step-by-step explanation:

5 0

3 years ago

Find the surface area of each figure. Round to the nearest tenth if necessary.​

Find the surface area of each figure. Round to the nearest tenth if necessary.