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Anastaziya [24]
4 years ago
9

Construct a truth table for (p ˄ ⌐q) ↔ (p ˄ q). Explain how this truth table shows whether this statement is a tautology, and co

ntradiction (absurdity), or a contingency.

Mathematics
1 answer:
lara [203]4 years ago
4 0

It's contingency, because for some values of p and q it's true, and for the other values, it's false.

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A rectangular yard is X feet wide. The yard is 4 feet longer than it is wide. The perimeter P of yard is given by the equation P
vladimir2022 [97]

Given:

Width of the rectangular yard = x feet

Length of the rectangular yard = x+4 feet

Perimeter of the rectangular yard is:

P=4x+8

Perimeter of the yard = 84 feet

To find:

The length of the rectangular yard.

Solution:

Perimeter of the yard is 84 feet.

4x+8=84

4x=84-8

4x=76

Divide both sides by 4.

x=\dfrac{76}{4}

x=19

So, the width of the rectangular yard is 19 feet.

Length = 19+4

Length = 23

The length of the rectangular field is 23 feet.

Therefore, the correct option is B.

6 0
3 years ago
if the heart of an elephant , at rest beat an average of 1680 beats in 60 min . What is the rate in beats per minute
Tresset [83]
Use the rule of 3 simple 

1680 beats ------- 60 min.
x beats -------------- 1 min. 
----------------------------------
 x = (1680*1)/60 = 1680/60 = 168/6 = 28 beats per minute 

hope helped 
5 0
3 years ago
Toms weekly salary increased from $240 to $270. what was the percent of change?
8_murik_8 [283]

Answer:

12.5%

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks
Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀  we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size  n = 15     n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean   μ₀  = 24,5

Sample mean   μ  =  25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis      H₀                             μ  =  μ₀

Alternative Hypothesis     Hₐ                  μ  >  μ₀

t (c) =  ?

We will define CI = 95 %  then   α = 5 %   α = 0,05    α/2 =  0,025

n = 15     then degree of freedom    df = 14

From t-student table  we get:  t(c) = 2,1448

And  t(s)

t(s) = ( μ  -  μ₀  ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare   t(c)   and  t(s)

t(c)  =  2,1448         t(s)  = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀   we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

8 0
3 years ago
SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!
Sloan [31]

ANSWER

P-G = 11{w}^{4} - 5 {w}^{2}  {z}^{2}   - 4z^{4}

EXPLANATION

The given polynomials are:

G =  - 3 {w}^{4}  + 2 {w}^{2}  {z}^{2}  + 5z^{4}

and

P = 8 {w}^{4}   - 3{w}^{2}  {z}^{2}  + z^{4}

We want to subtract these two polynomials.

P-G = 8 {w}^{4}   - 3{w}^{2}  {z}^{2}  + z^{4}  - (- 3 {w}^{4}  + 2 {w}^{2}  {z}^{2}  + 5z^{4} )

Expand the right hand side to obtain:

P-G = 8 {w}^{4}   - 3{w}^{2}  {z}^{2}  + z^{4}   + 3 {w}^{4}   -  2 {w}^{2}  {z}^{2}   -  5z^{4}

Group the similar terms:

P-G = 8 {w}^{4}  + 3 {w}^{4}- 3{w}^{2}  {z}^{2} -  2 {w}^{2}  {z}^{2}   -  5z^{4} + z^{4}

Combine the similar terms to get:

P-G = 11{w}^{4} - 5 {w}^{2}  {z}^{2}   - 4z^{4}

7 0
3 years ago
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