All categories

3 years ago

Please help me idk this

Mathematics

2 answers:

vovangra [49]3 years ago

8 0

Answer:

2.815

Step-by-step explanation:

8090 [49]3 years ago

3 0

Answer:

2.815

Step-by-step explanation:

after the decimal the order is tenths hundreds thousands... so on and so forth.

You might be interested in

The product of 86 and the depth of the river

aniked [119]

Answer:

Step-by-step explanation:

Are you trying to find a variable expression? the product of 86 means multiplication so 86*n or 86n. Other than that I dont understand the question.

4 0

2 years ago

An interior automotive supplier places several electrical wires in a harness.Apull test measures the force required to pull spli

oksano4ka [1.4K]

Answer:

a) For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assumin Normal distribution

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical dsitribution

The result is on the figure attached.

b) For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

Step-by-step explanation:

For this case we have the following data:

28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4

The quantile-quantile or q-q plot is a graphical procedure in order to check the validity of a distributional assumption for a data set. We just need to calculate "the theoretically expected value for each data point based on the distribution in question".

If the values are asusted to the assumed distribution, we will see that "the points on the q-q plot will fall approximately on a straight line"

For this case our distribution assumed is normal.

Part a

For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assuming Normal distribution (0,1)

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical distribution

The result is on the figure attached.

Part b

For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

4 0

3 years ago

Suppose an airline policy states that all baggage must be box shaped with a sum of length, width, and height not exceeding 114

NISA [10]

Answer:

Step-by-step explanation:

Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.

The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.

Substituting 114 - 2s for h in the volume formula, we obtain:

V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)

This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:

----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2

Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.

Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in

and the volume is V = s^2(h) = 46,298.25 in^3

7 0

3 years ago

Please help! Find the area and the perimeter of the shaded regions below. Give your answer as a completely simplified exact valu

Nadya [2.5K]

Answer:

$The \ area \ of \ the \ figure = 4\dfrac{1}{2} \cdot \pi + 18 \ cm^2$

Step-by-step explanation:

Given that the figure is made up of portion of a square and a semicircle, we have;

BC ≅ AB = 6 cm

The area of semicircle BC with radius BC/2 = 3 is 1/2×π×r² = 1/2×π×3² = 4.5·π cm²

Triangle ABC = 1/2 × Area of square from which ABC is cut

The area of triangle ABC = 1/2×Base ×Height = 1/2×AB×BC = 1/2×6×6 = 18 cm²

The area of the figure = The area of semicircle BC + The area of triangle ABC

The area of the figure = 4.5·π cm² + 18 cm² = $\dfrac{9 \cdot \pi +36}{2} \ cm^2$ .

5 0

3 years ago

How do I do this Q11

SashulF [63]

Look at one of the vertices of the heptagon where two squares meet. The angles within the squares are both of measure 90 degrees, so together they make up 180 degrees.

All the angles at one vertex must clearly add up to 360 degrees. If the angles from the squares contribute a total of 180 degrees, then the two remaining angles (the interior angle of the heptagon and the marked angle) must also be supplementary and add to 180 degrees. This means we can treat the marked angles as exterior angles to the corresponding interior angle.

Finally, we know that for any convex polygon, the exterior angles (the angles that supplement the interior angles of the polygon) all add to 360 degrees (recall the exterior angle sum theorem). This means all the marked angles sum to 360 degrees as well, so the answer is B.

5 0

3 years ago