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Dmitry_Shevchenko [17]
3 years ago
7

2-1 whats the answer

Mathematics
2 answers:
Marysya12 [62]3 years ago
8 0

Answer:

<h2>2-1=1</h2><h3><em>hope it helps you have a great day keep smiling be happy stay safe</em></h3><h2 />
vaieri [72.5K]3 years ago
4 0

Answer:

2-1 is 1

Step-by-step explanation:

hope that it is helpful yo you

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Which quadratic function has a wider graph than y=2x^2?
ioda

so all those quadratics have their parent y = x², which has the graph of a "bowl".

and depending on the coefficient, the bowl is wider or not.

the larger the coefficient, the narrower the bowl, the smaller the coefficient, the wider the bowl.

well, 1/2 is smaller then 2, thus y = (1/2)x² is wider than y = 2x².

7 0
3 years ago
Find the inverse f(x)=6x+7
Lorico [155]

Answer:

f^{-1} (x) = \frac{x-7}{6}

Step-by-step explanation:

let y = f(x) , then rearrange making x the subject

y = 6x + 7 ( subtract 7 from both sides )

y - 7 = 6x ( divide both sides by 6 )

\frac{y-7}{6} = x

Change y back into terms of x with x = f^{-1} (x) , then

f^{-1} (x) = \frac{x-7}{6}

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3 years ago
Find the area under the standard normal probability distribution between the following pairs of​ z-scores. a. z=0 and z=3.00 e.
prohojiy [21]

Answer:

a. P(0 < z < 3.00) =  0.4987

b. P(0 < z < 1.00) =  0.3414

c. P(0 < z < 2.00) = 0.4773

d. P(0 < z < 0.79) = 0.2852

e. P(-3.00 < z < 0) = 0.4987

f. P(-1.00 < z < 0) = 0.3414

g. P(-1.58 < z < 0) = 0.4429

h. P(-0.79 < z < 0) = 0.2852

Step-by-step explanation:

Find the area under the standard normal probability distribution between the following pairs of​ z-scores.

a. z=0 and z=3.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 3.00) = 0.9987

Thus;

P(0 < z < 3.00) = 0.9987 - 0.5

P(0 < z < 3.00) =  0.4987

b. b. z=0 and z=1.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 1.00) = 0.8414

Thus;

P(0 < z < 1.00) = 0.8414 - 0.5

P(0 < z < 1.00) =  0.3414

c. z=0 and z=2.00

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 2.00) = 0.9773

Thus;

P(0 < z < 2.00) = 0.9773 - 0.5

P(0 < z < 2.00) = 0.4773

d.  z=0 and z=0.79

From the standard normal distribution tables,

P(Z< 0) = 0.5  and P (Z< 0.79) = 0.7852

Thus;

P(0 < z < 0.79) = 0.7852- 0.5

P(0 < z < 0.79) = 0.2852

e. z=−3.00 and z=0

From the standard normal distribution tables,

P(Z< -3.00) = 0.0014  and P(Z< 0) = 0.5

Thus;

P(-3.00 < z < 0 ) = 0.5 - 0.0013

P(-3.00 < z < 0) = 0.4987

f. z=−1.00 and z=0

From the standard normal distribution tables,

P(Z< -1.00) = 0.1587  and P(Z< 0) = 0.5

Thus;

P(-1.00 < z < 0 ) = 0.5 -  0.1586

P(-1.00 < z < 0) = 0.3414

g. z=−1.58 and z=0

From the standard normal distribution tables,

P(Z< -1.58) = 0.0571  and P(Z< 0) = 0.5

Thus;

P(-1.58 < z < 0 ) = 0.5 -  0.0571

P(-1.58 < z < 0) = 0.4429

h. z=−0.79 and z=0

From the standard normal distribution tables,

P(Z< -0.79) = 0.2148  and P(Z< 0) = 0.5

Thus;

P(-0.79 < z < 0 ) = 0.5 -  0.2148

P(-0.79 < z < 0) = 0.2852

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Oduvanchick [21]

Answer:

the answer is 3

Step-by-step explanation:

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8 0
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Answer:

$520.83

Step-by-step explanation:

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