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4 days ago

Circle D circumscribes ABC and ABE. Which statements about the triangles are true? Statement I: The perpendicular bisectors of A

BC intersect at the same point as those of ABE. Statement II: The distance from C to D is the same as the distance from D to E. Statement III: bisects CDE. Statement IV: The angle bisectors of ABC intersect at the same point as those of ABE. A. I only B. I and II C. II and IV D. I and III E. III and IV

Mathematics

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You are given a rectangular sheet of cardboard that measures 11 in. by 8.5 in. (see the diagram below). A small square of the sa

AnnZ [12381]

Answer:

Terrence's

Step-by-step explanation:

The cut length of the square corresponds to the box height.

1. a

In Anya's approach: 8.5 - 1.5 = 7, 11 - 1.5 = 9.5; with the height being 1.5, the volume calculates as height x length x width = 1.5 x 9.5 x 7 = 99.75 square inches.

Terrence's method: 8.5 - 3 = 5.5, 11 - 3 = 8. Volume = 5.5 x 8 x 3 = 132 square inches. Clearly, 99.75 < 132 square inches, indicating Terrence's approach would yield a larger volume.

1. b

The dimensions of the box rely on the length, width, and height of the cardboard cutout, thus varying cutting measurements for the same cardboard size can produce different box dimensions.

2. Squares are cut from each corner, thus the combined size of the two squares on the cardboard must be less than the shorter cardboard side. The shorter side is 8.5 inches, which divided by 2 results in 4.25 inches, therefore squares cannot exceed this size. If squares are taken out measuring 4.25 inches, it will leave a strip 2.5 inches wide that cannot form a box.

To cut out squares of 5 inches, how you position them can cause overlap or extend beyond the cardboard. While 5 + 5 equals ten, making the longer side of 11 inches adequate, the 8.5 inches side allows zero room for the 10 inches.

7 0

3 months ago

The figure shows a kite inside a rectangle. Which expression represents the area of the shaded region? 2x2 4x2 6x2 8x2

lawyer [12517]

The area calculation for the shaded section, as seen in the attached diagram, involves subtracting the area of the kite from that of the rectangle. The area of the rectangle is calculated as (3x+x)*(x+x), which simplifies to 4x*2x, equating to 8x². The area of the kite is determined using the formula (1/2)*[d1*d2], where d1 and d2 represent the diagonals, specifically d1=4x and d2=2x. Therefore, the area of the kite becomes (1/2)*[4x*2x], leading to 4x². Consequently, the area of the shaded region can be computed as 8x²-4x², resulting in 4x². Thus, the solution is 4x².

5 0

3 months ago

Read 2 more answers

A study1 conducted in July 2015 examines smartphone ownership by US adults. A random sample of 2001 people were surveyed, and th

Inessa [12570]

Answer:

a) Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

b) $z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679$

c) $z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58$

d) In this scenario, we notice that $\hat p_1 > \hat p_2$ thus the conclusion for this case would indicate

Step-by-step explanation:

Information provided

$X_{1}=688$ denote the number of men possessing smartphones

$X_{2}=671$ signify the number of women possessing smartphones

$n_{1}=989$ group of men sampled

$n_{2}=1012$ group of women sampled

$p_{1}=\frac{688}{989}=0.696$ symbolize the proportion of men with smartphones

$p_{2}=\frac{671}{1012}=0.663$ symbolize the proportion of women with smartphones

$\hat p$ denote the pooled estimate of p

z would denote the test statistic

$p_v$ signify the value

Part a

The objective is to evaluate if there is a disparity in smartphone ownership between men and women; the hypothesis statements would be:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

Part b

The statistic relevant to this case is expressed as:

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$ (1)

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{688+671}{989+1012}=0.679$

Part c

By substituting the provided information, we find:

$z=\frac{0.696-0.663}{\sqrt{0.679(1-0.679)(\frac{1}{989}+\frac{1}{1012})}}=1.58$

Part d

In this instance, it is evident that $\hat p_1 > \hat p_2$ thus the conclusion for this case would seem

4 0

3 months ago

Last year a banquet hall charged $65 per person to attend a function. The function C(p) = 65p represents the total cost, C, to r

Svet_ta [12734]

Answer:

Domain = [0, 50]

Range = [0, 3250]

Step-by-step explanation:

A function illustrates the relationship between two variables (independent and dependent). The independent variable depends on nothing else, serving as the function's input, while the dependent variable relies on the independent variable, acting as the output.

The domain of a function includes all possible input variables (independent variable), and the range encompasses all potential output variables (dependent variables).

For the function C(p) = 65p, p represents the independent variable and C(p) constitutes the dependent variable.

As the hall accommodates a maximum of 50 individuals, the domain of this function is defined as [0, 50]

C(0) results in 65(0) = 0 and C(50) results in 65(50) = 3250

Thus, the range of this function is [0, 3250]

8 0

2 months ago

In triangle $abc$, $m\angle a = 90^\circ$, $m\angle b = 75^\circ$, and $bc = \sqrt {3}$ units. what is the area of triangle $abc

Inessa [12570]

Given the specified angles, one is 90 degrees, indicating that the triangle is a right triangle. With provided angles and one side representing the hypotenuse (the longest side), the area is determined using the formula: Area = 1/2 * base * height. Let's compute the height and base:

From sin 75, we derive height = 1.67.
And from cos 75, we obtain base = 0.45.

Calculating area gives us Area = (0.45 * 1.67) / 2, resulting in 0.37 square units.

Thus, the triangle's area is approximately 0.37 square units.

3 0

3 months ago