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mr_godi [17]
3 years ago
13

A circuit overloads at 1800 watts of electricity you plug in a toaster that uses 800 watts of electricity write and solve an ine

quality that represents how many watts you can add to the circuit without overloading it.
Mathematics
1 answer:
dangina [55]3 years ago
4 0

Answer:

x ≤ 1000 W

Step-by-step explanation:

Since the circuit becomes overloaded when the load exceeds 1800W and the toaster uses 800 W.

Let the additional watts be x

Therefore;

800 + x ≤ 1800

x ≤ 1800 -800

x ≤ 1000 W

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3 years ago
Dilate the trapezoid using center (-3,4) and scale factor 1/2.
pochemuha

The coordinates of the vertices of the image of the trapezoid are given as;

A'(x, y) = (- 4, 1), B'(x, y) = (- 2, 1), C'(x, y) = (- 5 / 2, 5 / 2) , D'(x, y) = (- 7 / 2, 5 / 2).

<h3>How to find the image of a trapezoid by dilation?</h3>

In this question we have a representation of a trapezoid, whose image has to be generated by a kind of rigid transformation known as dilation, whose equation is described :

P'(x, y) = O(x, y) + k · [P(x, y) - O(x, y)]

Where O(x, y) - Center of dilation

k - Scale factor

And P(x, y) - Coordinates of the original point, P'(x, y) - Coordinates of the resulting point.

Since k = 1 / 2, A(x, y) = (- 5, - 2), B(x, y) = (- 1, - 2), C(x, y) = (- 2, 1), D(x, y) = (- 4, 1), O(x, y) = (- 3, 4),

Therefore, the coordinates of the vertices of the image are:

Point A'

A'(x, y) = O(x, y) + k · [A(x, y) - O(x, y)]

A'(x, y) = (- 3, 4) + (1 / 2) [(- 5, - 2) - (- 3, 4)]

A'(x, y) = (- 3, 4) + (1 / 2)  (- 2, - 6)

A'(x, y) = (- 3, 4) + (- 1, - 3)

A'(x, y) = (- 4, 1)

Point B';

B'(x, y) = O(x, y) + k [B(x, y) - O(x, y)]

B'(x, y) = (- 3, 4) + (1 / 2) [(- 1, - 2) - (- 3, 4)]

B'(x, y) = (- 3, 4) + (1 / 2)  (2, - 6)

B'(x, y) = (- 3, 4) + (1, - 3)

B'(x, y) = (- 2, 1)

Point C';

C'(x, y) = O(x, y) + k · [C(x, y) - O(x, y)]

C'(x, y) = (- 3, 4) + (1 / 2)  [(- 2, 1) - (- 3, 4)]

C'(x, y) = (- 3, 4) + (1 / 2) (1, - 3)

C'(x, y) = (- 3, 4) + (1 / 2, - 3 / 2)

C'(x, y) = (- 5 / 2, 5 / 2)

Point D'

D'(x, y) = O(x, y) + k  [D(x, y) - O(x, y)]

D'(x, y) = (- 3, 4) + (1 / 2) [(- 4, 1) - (- 3, 4)]

D'(x, y) = (- 3, 4) + (1 / 2) (- 1, - 3)

D'(x, y) = (- 3, 4) + (- 1 / 2, - 3 / 2)

D'(x, y) = (- 7 / 2, 5 / 2)

To learn more on dilations:

brainly.com/question/13176891

#SPJ1

3 0
1 year ago
In an experiment to determine whether there is a systematic difference between the weights obtained with two different mass bala
Maslowich

Answer:

H0: μd=0 Ha: μd≠0

t= 0.07607

On the basis of this we conclude that the mean weight differs between the two balances.

Step-by-step explanation:

The null and alternative hypotheses as

H0: μd=0 Ha: μd≠0

Significance level is set at ∝= 0.05

The critical region is t ( base alpha by 2 with df=5) ≥ ± 2.571

The test statistic under H0 is

t = d/ sd/ √n

Which has t distribution with n-1 degrees of freedom

Specimen        A               B           d = a - b         d²

1                     13.76        13.74         0.02           0.004

2                    12.47        12.45          0.02         0.004

3                    10.09        10.08           0.01        0.001

4                       8.91       8.92          -0.01          0.001

5                     13.57      13.54           0.03        0.009

<u>6                     12.74      12.75          -0.01        0.001</u>

<u>∑                                                      0.06         0.0173</u>

d`= ∑d/n= 0.006/6= 0.001

sd²= 1/6( 0.0173- 0.006²/6) = 1/6 ( 0.017294) = 0.002882

sd= 0.05368

t= 0.001/ 0.05368/ √6

t= 0.18629/2.449

t= 0.07607

Since the calculated value of t= 0.07607 does not falls in the rejection region we therefore accept the null hypothesis at 5 % significance level . On the basis of this we conclude that the mean weight differs between the two balances.

3 0
3 years ago
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