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torisob [31]
2 years ago
7

3(9z+4) > 35z - 4 What is z

Mathematics
1 answer:
attashe74 [19]2 years ago
4 0

3(9z+4) > 35z - 4

27z + 12 > 35z - 4

12 + 4 > 35z - 27z

16 > 8z

\frac{16}{8}  > z

2 > z

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Use the net to compute the surface area of the three-dimensional figure.
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9514 1404 393

Answer:

  382 square units

Step-by-step explanation:

The central four rectangles down the middle of the net are 9 units wide, and alternate between 8 and 7 units high. Then the area of those four rectangles is ...

  9(8+7+8+7) = 270 . . . square units

The rectangles making up the two left and right "wings" of the net are 8 units high and 7 units wide, so have a total area of ...

  2×(8)(7) = 112 . . . square units

Then the area of the figure computed from the net is ...

  270 +112 = 382 . . . square units

__

<em>Additional comment</em>

You can reject the first two answer choices immediately, because they are odd. Each face will have an area that is the product of integers, so will be an integer. There are two faces of each size, so <em>the total area of this figure must be an even number</em>.

You may recognize that the dimensions are 8, 8+1, 8-1. Then the area is roughly that of a cube with dimensions of 8: 6×8² = 384. If you use these values (8, 8+1, 8-1) in the area formula, you find the area is actually 384-2 = 382. That area formula is A = 2(LW +H(L+W)).

3 0
3 years ago
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According to a recent​ survey, the population distribution of number of years of education for​ self-employed individuals in a c
creativ13 [48]

Answer:

a) X: number of years of education

b) Sample mean = 13.5, Sample standard deviation = 0.4

c) Sample mean = 13.5, Sample standard deviation = 0.2

d) Decrease the sample standard deviation

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 13.5 years

Standard deviation,σ = 2.8 years

a) random variable X

X: number of years of education

Central limit theorem:

If large random samples are drawn from population with mean \mu and standard deviation \sigma, then the distribution of sample mean will be normally distributed with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

b) mean and the standard for a random sample of size 49

\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{49}} = 0.4

c) mean and the standard for a random sample of size 196

\mu_{\bar{x}} = \mu = 13.5\\\\\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2.8}{\sqrt{196}} = 0.2

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