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Allisa [31]
1 year ago
15

Let p: x=4 let q:y=-2

Mathematics
1 answer:
ICE Princess25 [194]1 year ago
6 0

Question:

Let p: x = 4 Let q: y = −2 Which represents "If x = 4, then y = −2”? p ∨ q p ∧ q p → q p ↔ q

Answer:

Option c: p → q

Explanation:

Option a: p ∨ q

p ∨ q means that either p is true or q is true or both are true.

Hence, Option a is not the correct answer.

Option b: p ∧ q

p ∧ q means that both p and q are true.

Hence, Option b is not the correct answer.

Option c: p → q

p → q means that if p is true, then q is also true.

Hence, Option c is the correct answer.

Option d: p ↔ q

p ↔ q means that if p is true, then q is true. If q is true, then p is true.

Hence, Option d is not the correct answer.

Thus, p → q is the correct option.

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A process has been developed that can transform ordinary iron into a kind of super iron called metallic glass. Metallic glass is
beks73 [17]

Answer:

Step-by-step explanation:

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6 0
1 year ago
which represents the solution(s) of the system of equations, y=x^2-2x-15 and y=8x-40? determine the solution set algebraically
Harrizon [31]
y=x^2-2x-15 (1) y=8x-40 (2)

8x-40=x^2-2x-15
X^2+10x-25=0
(x-5)^2
×=5 y=0
7 0
2 years ago
A process manufactures ball bearings with diameters that are normally distributed with mean 25.1 mm and standard deviation 0.08
marta [7]

Answer:

(a) The proportion of the diameters are less than 25.0 mm is 0.1056.

(b) The 10th percentile of the diameters is 24.99 mm.

(c) The ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d) The proportion of the ball bearings meeting the specification is 0.8881.

Step-by-step explanation:

Let <em>X</em> = diameters of ball bearings.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 25.1 mm and standard deviation, <em>σ</em> = 0.08 mm.

To compute the probability of a Normally distributed random variable we need to first convert the raw scores to <em>z</em>-scores as follows:

<em>z</em> = (X - μ) ÷ σ

(a)

Compute the probability of <em>X</em> < 25.0 mm as follows:

P (X < 25.0) = P ((X - μ)/σ < (25.0-25.1)/0.08)

                    = P (Z < -1.25)

                    = 1 - P (Z < 1.25)

                    = 1 - 0.8944

                    = 0.1056

*Use a <em>z</em>-table for the probability.

Thus, the proportion of the diameters are less than 25.0 mm is 0.1056.

(b)

The 10th percentile implies that, P (X < x) = 0.10.

Compute the 10th percentile of the diameters as follows:

P (X < x) = 0.10

P ((X - μ)/σ < (x-25.1)/0.08) = 0.10

P (Z < z) = 0.10

<em>z</em> = -1.282

The value of <em>x</em> is:

z = (x - 25.1)/0.08

-1.282 = (x - 25.1)/0.08

x = 25.1 - (1.282 × 0.08)

  = 24.99744

  ≈ 24.99

Thus, the 10th percentile of the diameters is 24.99 mm.

(c)

Compute the value of P (X < 25.2) as follows:

P (X < 25.2) = P ((X - μ)/σ < (25.2-25.1)/0.08)

                    = P (Z < 1.25)

                    = 0.8944

                    ≈ 0.84

*Use a <em>z</em>-table for the probability.

Thus, the ball bearing that has a diameter of 25.2 mm is at the 84th percentile.

(d)

Compute the value of P (25.0 < X < 25.3) as follows:

P (25.0 < X < 25.3) = P ((25.0-25.1)/0.08 < (X - μ)/σ < (25.3-25.1)/0.08)

                    = P (-1.25 < Z < 2.50)

                    = P (Z < 2.50) - P (Z < -1.25)

                    = 0.99379 - 0.10565

                    = 0.88814

                    ≈ 0.8881

*Use a <em>z</em>-table for the probability.

Thus, the proportion of the ball bearings meeting the specification is 0.8881.

4 0
1 year ago
The constant-pressure specific heat of air at 25°C is 1.005 kJ/kg. °C. Express this value in kJ/kg.K, J/g.°C, kcal/ kg. °C, and
Mice21 [21]

Answer:

In kJ/kg.K - 1.005  kJ/kg degrees Kalvin.

In  J/g.°C  -  1.005 J/g °C

In kcal/ kg °C  0.240 kcal/kg °C

In Btu/lbm-°F   0.240 Btu/lbm degree F

Step-by-step explanation:

given data:

specific heat of air = 1.005 kJ/kg °C

In kJ/kg.K

1.005 kJ./kg °C = 1.005 kJ/kg degrees Kelvin.

In  J/g.°C

1.005 kJ./kg °C \times 1000/1kJ \times (1kg/1000 g) = 1.005J/g °C

In kcal/ kg °C

1.005 kJ./kg °C \times \frac{1 kcal}{4.190 kJ} = 0.240 kcal/kg °C

For   kJ/kg. °C to Btu/lbm-°F  

Need to convert by taking following conversion ,From kJ to Btu, from kg to lbm and from degrees C to F.

1.005 kJ./kg °C \times \frac{1 Btu}{1.055 kJ} \times \frac{0.453 kg}{1 lbm} \times \frac{5/9 degree C}{1 degree F} = 0.240 Btu/ lbm / degree F

1.005 kJ/kg C =  0.240 Btu/lbm degree F

8 0
2 years ago
What is the solution set for - 4x + 10 = 5(x + 11)? HELPPPPPPP​
Scorpion4ik [409]

Answer:

-5

Step-by-step explanation:

-4x +10 =5(x +11)

-4x +10 =5x +55

-4x - 5x =55 - 10

-9x =45

x=-45 :9

x=-5

7 0
1 year ago
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