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dlinn [17]
3 years ago
11

Write a grammar for a language whose sentences start with an even and non-zero number of x’s, end with an odd number of z’s, and

have a number of y’s in between equal to the total number of x’s and z’s in the sentence. For example, xxyyyyyzzz is a valid sentence.
Engineering
1 answer:
sveta [45]3 years ago
3 0

Answer:

GRAMMAR

S -> AB

A -> xxAyy | xxyy

B -> yyBzz | yz

EXPLANATION

A is for even number of x's followed by that number of y's

B is for odd number of y's followed by that number of z's

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An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc
Vadim26 [7]

Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

Sample mean : \overline{x}=2.45

Standard deviation : \sigma=0.07

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, z=\dfrac{x-\mu}{\sigma}, the z-value corresponds to 2.39 will be :-

z=\dfrac{2.39-2.45}{0.07}\approx-0.86

z-value corresponds to 2.60 will be :-

z=\dfrac{2.60-2.45}{0.07}\approx2.14

Using the standard normal table for z, we have

P-value = P(-0.86

=P(z

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

7 0
3 years ago
Sharon is designing a house in an area that receives a lot of rainfall all year. Which material should she use to stick the wood
kakasveta [241]

Explanation:

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3 years ago
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Airida [17]
I think downwards as that's how most saw's work.
4 0
3 years ago
When the psychologist simply records the relationship between two variables...
Wewaii [24]
When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.

The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.
8 0
3 years ago
When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b
laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

\tau_{ave} = \frac{T}{2tA_{m}}

A_m = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, A_m = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

\tau_{ave} = \frac{T}{2tA_{m}}

Plugging in then values, gives;

\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa

The shear stress will be 80,000,000 Pa or 80 MPa.

7 0
3 years ago
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