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Kay [80]
4 years ago
6

PLEASE HELP I FAILED THIS SO MANY TIMES ????

Mathematics
1 answer:
Likurg_2 [28]4 years ago
7 0
So it would be a,b because it like that
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What is -7(x-2)=1=15-7x
Soloha48 [4]

Answer:

0

Step-by-step explanation:

Simplifying

-7(x + -2) + 1 = 15 + -7x

Reorder the terms:

-7(-2 + x) + 1 = 15 + -7x

(-2 * -7 + x * -7) + 1 = 15 + -7x

(14 + -7x) + 1 = 15 + -7x

Reorder the terms:

14 + 1 + -7x = 15 + -7x

Combine like terms: 14 + 1 = 15

15 + -7x = 15 + -7x

Add '-15' to each side of the equation.

15 + -15 + -7x = 15 + -15 + -7x

Combine like terms: 15 + -15 = 0

0 + -7x = 15 + -15 + -7x

-7x = 15 + -15 + -7x

Combine like terms: 15 + -15 = 0

-7x = 0 + -7x

-7x = -7x

Add '7x' to each side of the equation.

-7x + 7x = -7x + 7x

Combine like terms: -7x + 7x = 0

0 = -7x + 7x

Combine like terms: -7x + 7x = 0

0 = 0

Solving

0 = 0

Couldn't find a variable to solve for.

This equation is an identity, all real numbers are solutions.

3 0
3 years ago
Please help me with this
docker41 [41]

Answer: y=12.287

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
A movie theater charges $10 for adult tickets and $7 for child tickets. On a particular day, they made $1446 in ticket sales. Th
luda_lava [24]
X = adults
Y = child

X+y=168
X=168-y

$10x + $7y = $1446

10(168-y) + 7y = 1446
1680 - 10y + 7y = 1446
-3 y = -234
Y = 78

X = 168-78
X = 90

$10*90 + $7*78 =
$900 + $546 = $1446
6 0
3 years ago
Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. ​(a) Descr
Daniel [21]

Answer:

(a) The sampling distribution of\overline{X} = Population mean = 79

(b)  P ( \overline{X} greater than 81.2 ) =  0.1357

(c) P (\overline{X} less than or equals 74.4 ) = .0107

(d) P (77.6 less than \overline{X} less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean (\nu) = 79

Standard deviation (\sigma ) = 18

​(a) Describe the sampling distribution of \overline{X}

For large sample using central limit theorem

the sampling distribution of\overline{X} = Population mean = 79

​(b) What is Upper P ( \overline{X} greater than 81.2 )​ =

P(\overline{X}> 81.2)  = P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})

                    =  P(Z> 1.1)

                    = 1 - P(Z<   1.1)

                    = 1 - .8643 =

                    = 0.1357

(c) What is Upper P (\overline{X} less than or equals 74.4 ) =

P(\overline{X}\leq  74.4) = P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq  \frac{74.4- 79}{\frac{18}{\sqrt{81}}})

                    = P(Z\leq  -2.3)

                    = .0107

​(d) What is Upper P (77.6 less than \overline{X} less than 83.2 ) =

P(77.6< \overline{X}<   83.2) = P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq  \frac{83.2- 79}{\frac{18}{\sqrt{81}}})

                                = P(- 0.7< Z<   2.1)

                                 = (Z<   2.1) - (Z<   -0.7)

                                  = 0.9821 - .2420

                                   = 0.7401

3 0
3 years ago
I don’t need a explanation, I just need to know if these 3 graphs are growth or decay. This needs to be in within a hour
riadik2000 [5.3K]

Hey there!

The answer to your question is:

Graph A represents a growth.

Graph B represents a decay.

Graph C represents a growth.

I will give an explanation any way.

Graph A has a positive slope - a growth.

Graph B has a negative slope - a decay.

Graph C has a positive slope - a growth.

Good luck on your assignment! Hope it helps, and have a great day!

7 0
3 years ago
Read 2 more answers
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