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3 years ago

Howwww?????? Answer???

Mathematics

2 answers:

Anika [276]3 years ago

8 0

8x - 12 = 6x + 2

8x - 12 + 12 = 6x + 2 + 12

8x = 6x + 14

8x - 6x = 6x + 14 - 6x

2x = 14

2x/2 = 14/2

X = 7

Sever21 [200]3 years ago

4 0

Answer:

x=7

Step-by-step explanation:

First, you need to expand the brackets.

8x-12=6x+2

Next, you need to get x on one side by subtracting 6x from both sides.

8x-12-6x=6x+2-6x (2x-12=2).

Now, add 12 to both sides of the equation

2x-12+12=2+12 (2x=14).

And finally, divide both sides by 2.

2x÷2=14÷2 (x=7)

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uv = 4

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4 years ago

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3 years ago

Anyone know the answer???

alisha [4.7K]

Hello from MrBillDoesMath!

Answer:

k = 10

Discussion:

angle MYK = 180 and angle MYX = 180 => angle ZYX = 180 - 115 - 65

Trialing XYZ has 180 degrees so

180 = (4k + 5) + (6k+10) + 65 => combine like terms

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3 years ago

Items are inspected for flaws by two quality inspectors. Both inspectors inspect every item and the probability that an item has

Genrish500 [490]

Answer:

a)0.976

b)0.00926

c)0.2402

d)0.35

Step-by-step explanation:

Let $X_i$ be an item passed by inspector i

Let Y be the event that there is a fault in an item

The probability that an item has a flaw is 0.1 i.e. P(Y)=0.1

If a flaw is present ,it will be detected by the first inspector with probability 0.92 i.e. $P(\bar{X_1}|Y)=0.92$

So, $P(X_1|Y)=1-0.92=0.08$

If a flaw is present ,it will be detected by the second inspector with probability 0.7 i.e. $P(\bar{X_2}|Y)=0.7$

So, $P(X_2|Y)=1-0.7=0.3$

If an item does not have a flaw, it will be passed by the first inspector with probability 0.95 i.e. $P(X_1|\bar{Y}) = 0.95$

So, $P(\bar{X_1}|\bar{Y}) = 1-0.95=0.05$

If an item does not have a flaw, it will be passed by the second inspector with probability 0.8 i.e. $P(X_2|\bar{Y}) = 0.8$

So, $P(\bar{X_2}|\bar{Y}) = 1-0.8=0.2$

a)P(found by atleast one inspector | It has flaw )=1-P(found by none inspector | It has flow )

P(found by atleast one inspector | It has flaw )= $1-P(X_1|Y) P(X_2|Y)$

P(found by atleast one inspector | It has flaw )= $1-0.08 \times 0.3$

P(found by atleast one inspector | It has flaw )=0.976

Hence the probability that it will be found by at least one of the two inspectors if it has flaw is 0.976

b) $P(Y|X_1)=\frac{P(X_1|Y) P(Y)}{P(X_1|Y) P(Y)+P(X_1|\bar{Y}) P(\bar{Y})}$

$P(Y|X_1)=\frac{0.08 \times 0.1}{0.08 \times 0.1+0.95 \times 0.9}=0.00926$

C)P( two inspectors draw different conclusions on the same item)= $P(X_1 \cap \bar{X_2} \cap Y)+P(\bar{X_1} \cap X_2 \cap Y)+P(X_1 \cap \bar{X_2} \cap \bar{Y})+P(\bar{X_1} \cap X_2 \cap \bar{Y})$

P( two inspectors draw different conclusions on the same item)=0.2402

$P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2)}\\P(Y|(X_1 \cap X_2))=\frac{P(Y \cap X_1 \cap X_2)}{P(X_1 \cap X_2 \cap Y)+P(X_1 \cap X_2 \cap \bar{Y})}\\P(Y|(X_1 \cap X_2))=0.35$

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Dmitriy789 [7]

Looks like the two lines meet at (1,-1) so B is the answer.

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