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PtichkaEL [24]
2 years ago
6

How to find displacement from velocity.

Mathematics
1 answer:
Ipatiy [6.2K]2 years ago
5 0

Answer:

  • <em>The average velocity of the object is multiplied by the time traveled to find the displacement. The equation x = ½( v + u)t can be manipulated, as shown below, to find any one of the four values if the other three are known.</em>

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The Pythagorean theorem is a formula used in solving for missing side lengths of a(n)_______________ triangle.
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The third answer choice; right triangle.
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Evaluate the following expression 2+(3-2×2)×1
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2•3=9 2•2=4 2•2=4 9+4+4=17 the answer is 17
7 0
3 years ago
Increase £2193.67 by 8.5%
HACTEHA [7]

Answer:

£2380.23

Step-by-step explanation:

The increase will be 8.5% of £2193.76 added to £2193.76

That’s

8.5% of £2193.76

8.5% /100% x £2193.76

0.085 x £2193.76

£186.47

Increase = £2193.76 + £186.47

= £2380.23

6 0
3 years ago
A study was performed to investigate whether teens and adults had different habits when it comes to consuming meat-free meals. I
Elden [556K]

Answer:

H₀ should be rejected at CI 95% .

Then the proportion of adults with 95 % CI is bigger than the proportion of teen

Step-by-step explanation:

From adult sample:

n₂ = 2323

x₂ = 1601

p₂ = 1601 / 2323         p₂  = 0,689       or    p₂ = 68,9%

From teen sample:

n₁ = 875

x₁ = 555

p₁ = 555/ 875           p₁  = 0,634          or    p₁  = 63,4

Values of p₁  and  p₂   suggest that the proportion of adults consuming at least one meat free meal per week is bigger than teen proportion.

To either prove or reject the above statement we have to develop a difference of proportion test according to:

Hypothesis test:

Null Hypothesis                   H₀             p₁  =  p₂

Alternative Hypothesis       Hₐ             p₂  > p₁

So is a one-tail test to the right

We can establish a confidence interval of 95 % then  α = 5 %

or    α = 0,05

As the samples are big enough we will develop a z test

Then  z(c)  for α = 0,05   from z table is     z(c) = 1,64

To calculate z(s)

z(s) = ( p₂  -  p₁ ) / √p*q* ( 1/n₁  + 1/n₂ )

where p = ( x₁ + x₂ ) / n₁ + n₂       p = 555 + 1601 / 875 + 2323

p = 2156/3198       p = 0,674

and   q = 1 - 0,674        q = 0,326

z(s) = ( 0,689 - 0,634 ) / √0,674*0,326 ( 1/875 + 1 / 2323

z(s) = 0,055/ √ 0,2197 ( 0,00114 + 0,00043)

z(s) = 0,055/ √0,2197* 0,00157

z(s) = 0,055/ √ 3,45*10⁻⁴

z(s) = 0,055 / 1,85*10⁻²

z(s) = 5,5/1,85

z(s) = 2,97

Comparing  z(s) and z(c)     z(s) > z(c)

Then z(s) is in the rejection region we reject H₀.

We can claim that the proportion of adult eating at least one meat-free meal is bigger than the proportion of teen

6 0
3 years ago
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