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3 years ago

classify each scale factor as a contraction or an expansion Dragon drop the choices into the box to complete the table

Mathematics

2 answers:

TEA [102]3 years ago

8 0

Anything that is multiplied by a number over 1 will get bigger. any other number such as negatives or fractions/decimals will get smaller. 1.25 and 3 are expansions and -2,3/4, and 1/5 are contractions

dybincka [34]3 years ago

6 0

Answer: Contraction = $\dfrac{3}{4},\ \dfrac{1}{5}$

Expansion = 1.25, -3, -2

Step-by-step explanation:

Let k be the scale factor .

Expansion : |k|>1

Contraction : 0<|k|<1

Let's check the given scale factors :

1.25> 1, then this scale factor is an expansion.

3> 1, then this scale factor is an expansion.

|-2|=2>1, then this scale factor is an expansion.

$\dfrac{3}{4}=0.75$

& 0<0.75<1 , thus $\dfrac{3}{4}$ is an contraction.

$\dfrac{1}{5}=0.2$

& 0<0.2<1 , thus $\dfrac{1}{5}$ is an contraction.

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The axis of symmetry for a function in the form f(x)=x2+4x-5 is x=-2

Valentin [98]

Answer:

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2 years ago

Abatchof140semiconductorchipsisinspected by choosing a sample of 5 chips without replacement. Assume 10 of the chips do not conf

Mkey [24]

Answer:

Step-by-step explanation:

given that a batch of 140 semiconductor chips is inspected by choosing a sample of 5 chips without replacement.

10 chips do not conform to customer requirements.

i.e. out of 140 chips 10 are defective and 130 are good

a) 5 chips can be selected form 140 chips in 140C5 ways

= 416965528

b) For containing exactly one non conforming chip we have one selected from 10 and other 4 selected from 130

No of ways = $10C1 (130C4)\\=10(11358880)\\=113588800$

c) For atleast one nonconforming chip we can calculate as total - no non conforming

i.e. $416965528-130C5\\= 416965528-286243776\\=130721752$

4 0

3 years ago

A quality control technician works in a factory that produces computer monitors. Each day, she randomly selects monitors and tes

jeka94

Answer:

There is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 300

p = 5% = 0.05

Alpha, α = 0.05

Number of dead pixels , x = 24

First, we design the null and the alternate hypothesis

$H_{0}: p = 0.05\\H_A: p > 0.05$

This is a one-tailed(right) test.

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{24}{300} = 0.08$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.08-0.05}{\sqrt{\frac{0.05(1-0.05)}{300}}} = 2.384$

Now, we calculate the p-value from excel.

P-value = 0.00856

Since the p-value is smaller than the significance level, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.

Conclusion:

Thus, there is enough evidence to support the claim that the true proportion of monitors with dead pixels is greater than 5%.

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2 years ago

Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea

Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

y''(t) + p(t) y'(t) + q(t) y(t) = f(t)

so the first thing you need to do is divide both sides of your equation by t :

y'' + (2t - 1)/t y' - 2/t y = 7t

You're looking for a solution of the form

$y=y_1u_1+y_2u_2$

where

$u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

$u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

and W denotes the Wronskian determinant.

Compute the Wronskian:

$W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}$

Then

$u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t$

$u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}$

The general solution to the ODE is

$y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}$

which simplifies somewhat to

$\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}$

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2 years ago

12 students share 3 pizzas equally what fraction of the pizza each student gets

alekssr [168]

To find the solution, simply take 3 and divide it by 12. 3/12 as a fraction would be simplified down to 1/4 or 0.25. Therefore, each student would receive one quarter of a pizza.

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3 years ago