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2 years ago

9

How many sides does a dodecahedron have?

1 answer:

777dan777 [17]2 years ago

6 0

Answer:

twelve faces

Step-by-step explanation:

12 sides

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Identify the volume of the composite figure rounded to the nearest tenth. HELP PLEASE!!

nirvana33 [79]

Answer:

V = 115.3 ft³

Step-by-step explanation:

The left part of the figure shows a cube of side length 4.2 ft. The volume of a cube is V = s³, where s is the side length. Hence, the volume of this particular cube is V = (4.2 ft)³ = 74.088.

The volume of a pyramid is V = (1/3)(base area)(height).

Here V = (1/3)(4.2 ft)²(7 ft) = 41.16 ft³.

Summing up the two distinct areas, we get V = 41.16 ft³ + 74.088 ft³, or

V = 115.3 ft³ after rounding up to the nearest tenth.

3 0

2 years ago

Complete the general form equation of the parabola that passes through (4, -11) with vertex at : (2, -3).

fomenos

Answer:

Step-by-step explanation:

You're given a coordinate in the form of (x, y) and you're also given the vertex in the form of (h, k). We will use those in the vertex form of the equation

$y=a(x-h)^2+k$ and solve for a. Filling in:

$-11=a(4-2)^2-3$ which simplifies a bit to

$-11=a(2)^2-3$ and a bit more to

$-11=4a-3$ . Add 3 to both sides to get

-8 = 4a so

a = -2

The equation, then, is

$y=-2(x-2)^2-3$

7 0

2 years ago

Help please, would it be the answer choice C or D

UkoKoshka [18]

Answer:

probably C

Explanation:

5 0

3 years ago

#1. Graph the following equation.<br> Y=6/5x + 1

TEA [102]

Answer:

6x + 1

y= 1_ 6x

it will help you

5 0

2 years ago

Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. (a) Descri

OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

$\mu = 74$

and

$\sigma = 6$

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

$\mu = 74$

The standard error of the means becomes the standard deviation of the sampling distribution.

$\sigma_ { \bar X } = \frac{ \sigma}{ \sqrt{n} } \\ \sigma_ { \bar X } = \frac{ 6}{ \sqrt{36} } = 1$

Part b) We want to find

$P(\bar X \:>\:75.9)$

We need to convert to z-score.

$P(\bar X \:>\:75.9) = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: \frac{75.9 - 74}{1} ) \\ = P(z \:>\: 1.9) \\ = 0.0287$

Part c)

We want to find

$P(\bar X \: < \:71.95)$

We convert to z-score and use the normal distribution table to find the corresponding area.

$P(\bar X \: < \:71.95) = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: \frac{71.9 5- 74}{1} ) \\ = P(z \: < \: - 2.05) \\ = 0.0202$

Part d)

We want to find :

$P(73\:$

We convert to z-scores and again use the standard normal distribution table.

$P( \frac{73 - 74}{1} \:< \: z$

5 0

2 years ago