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3 years ago

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Find the solution set of this inequality. select the correct graph. |-4x+14|>6

1 answer:

lions [1.4K]3 years ago

5 0

Split the inequality into two:
(-4x+14)>6 & -(-4x+14)>6

solve:
(-4x+14)>6
-4x>-8
x<2
(symbol changes when ÷/× by negative for inequalities.)

-(-4x+14)>6
-4x+14<-6
-4x<-20
x>5

the answers are x<2 and x>5, so you would write your answer like this: (-∞,2) ∪ (5,∞)

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Triangle ABC is a scaled copy of triangle DEF. Side AB measures 12

Dima020 [189]

Answer:

1. The scale factor here is 1.5

2. The scale factor here is 2/3

Step-by-step explanation:

Here, we shall be dealing with scales of triangles.

we have two triangles;

ABC and DEF

longest sides are in the ratio;

12 : 8

1. What scale factor translates DEF to ABC?

The ratio of the length can be beaten down to 3:2

So therefore, we can see that by multiplying the sides of of DEF by 1.5, we can arrive at the sides of ABC

So the scale factor here is 1.5

2. This is like the other way round of what we have above.

By multiplying the sides of ABC by 2/3, we shall have the sides of DEF

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3 years ago

How do you find the slope of the line that goes<br> through the points (-5, 6) and (7,-6)?

strojnjashka [21]

Answer:

Step-by-step explanation:

$slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{-6-6}{7-[-5]}\\\\=\frac{-12}{7+5}\\\\=\frac{-12}{12}\\\\=-1$

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3 years ago

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Ivahew [28]

2665680 is the answer

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3 years ago

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It takes liam 56 minutes to lift 4 tyres<br> If he works for 42 minutes how many tyres would he lift

vagabundo [1.1K]

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Step-by-step explanation:

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3 years ago

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Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2

leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where $\lambda = \frac{10.2}{200} = 0.051$

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

$\frac{1}{4} \times 200 = 50$

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

$P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004$

c)The expected number of grams to be eaten before encountering the first fragments :

$E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 gram$ s

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3 years ago