3 years ago

PLEASE HELP i have no clue what the answer is

Mathematics

1 answer:

Nat2105 [25]3 years ago

5 0

Answer:

y=2/3x+19/3

Step-by-step explanation:

Use this equation y-y1=m(x-x1) and plug the numbers in

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Can someone please!!! Help me

timurjin [86]

The answer to this question is true

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3 years ago

HELP ASAP ASAP!! NO LINKS PLEASEEEE OR U WILL BE REPORTED, I NEED AN ANSWER WITH EXPLANATION AND WORK SHOWN OR EXPLAINED, WILL G

iogann1982 [59]

Graph these numbers and then connect at the end.

3. y (6,-3). x (2,-4) z (5,-7)

4. U (5,4). T (5,1) S (1,0) V(-3,4)

i hope these help.

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2 years ago

<img src="https://tex.z-dn.net/?f=%28%5Cfrac%7B8a%5E%7B-5%7Db%5E%7B4%7D%20%7D%7B12a%5E%7B-6%7Db%5E%7B-2%7D%20%7D%29" id="TexForm

Oxana [17]

Answer:

$→ \frac{8 {a}^{( - 5) } {b}^{(4)} }{12 {a}^{( - 6)} {b}^{( - 2)} } = \frac{2 {a}^{(6 - 5)}{b}^{(4 + 2)} }{3} = \boxed{ \frac{2}{3}a {b}^{6} }✓ \\$

<u>(2/3)ab⁶</u> is the right answer.

6 0

3 years ago

How to do 3(x-2)= -6

nasty-shy [4]

Answer:

3×x -3×2=-6

or,3x-6=-6

or,3x=-6+6

or,x=0/3

therefore=x =1 answer

8 0

3 years ago

Read 2 more answers

The test results for a class with 24 students are normally distributed with a mean of 82 and a standard deviation of 8.9. What i

alekssr [168]

Answer:

The value is $P( X > 80 ) = 0.58901$

Step-by-step explanation:

From the question we are told that

The sample size is n = 24

The mean is $\mu = 82$

The standard deviation is $\sigma = 8.9$

Generally the probability that a randomly chosen student earned a B or higher on the test (score of 80 or higher ) is mathematically represented as

$P( X > 80 ) = P( \frac{ X - \mu }{\sigma } > \frac{ 80 - 82 }{ 8.9 } )$

$\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

=> $P( X > 80 ) = P(Z > -0.225 )$

From the z table the area under the normal curve to the right corresponding to -0.225 is

$P(Z > -0.225 ) = 0.58901$

=> $P( X > 80 ) = 0.58901$

5 0

3 years ago

￼PLEASE HELP i have no clue what the answer is

PLEASE HELP i have no clue what the answer is