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erastovalidia [21]
3 years ago
14

Please quickly help!! I will Mark brainliest, What is the inverse of the function f(x) = x +3? Thank you!

Mathematics
2 answers:
Irina18 [472]3 years ago
8 0

Answer:

Mark me as Brainliest

fx^{-1} (x)=x+3

Step-by-step explanation:

To find the inverse of a function, just “trade” x and y and solve for the “new” y. The graph of the inverse is the reflection of the original function over the line y=x.

Rewrite the function

f(x)=x+3

as the equation, y=x+3

Trade x and y.

X=y+3

The symbol for the inverse if fx^{-1} (x)

fx^{-1} (x)=x+3

hichkok12 [17]3 years ago
7 0

Answer:

x-3

Step-by-step explanation:

hlo therre!!

<em>f(x)=x+3</em>

<em>let f(x) be y;</em>

<em>y=x+3</em>

<em>now; interchanging the value of x and y;</em>

<em>i.e. x=y+3</em>

<em>y=x-3</em>

<em>f(x)^-1 = x - 3...</em>

<em> this is the required inverse function..</em>

<em>thank u!! hope it helps!</em>

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Laryngeal cancer rates in smokers is 160.0 (per 100,000) and 25.0 (per 100,000) among nonsmokers. Among smokers, what percentage
Romashka [77]

Answer:

0.16%

Step-by-step explanation:

From the statement of the question;

Number of Laryngeal cancer due to smoking = 160

Population of smokers = 100,000

Hence the percentage of smokers liable to have Laryngeal cancer = 160/100000 ×100/1

=0.16%

Hence 0.16% of smokers are liable to Laryngeal cancer

4 0
3 years ago
A variable contains five categories. It is expected that data are uniformly distributed across these five categories. To test th
lubasha [3.4K]

Answer:

33.293 ± 0.01= 33.303 and 33.383,

Step-by-step explanation:

We first need to fit a normal distribution , but neither the mean  nor the standard deviation is given . We therefore estimate the sample mean and sample standard deviation  <em>s</em>. Using the data we find ∑fx=<u>378 </u>  and

<u>∑fx²=1344   </u> so that mean x` = 2.885 or 2.9   and standard deviation s =1.360

x        f                fx       x²         fx²

1       27             27        1          27

2       30           60         4          120

3       29           87         9          261

4       21            84         16        336

<u>5       24           120       25        600           </u>

<u>      ∑f=131      ∑fx=378             ∑fx²=1344   </u>

Mean = x`=<u> </u> ∑fx/ <u>  </u>∑f=  2.9

Standard Deviation = s= √∑fx²/∑f-(∑fx/∑f)²

                  s= √1344/131 - (378/131)²

                   s= √10.26-(2.9)²

                     s= √10.26- 8.41

                    s= √1.85= 1.360

Next we need to compute the expected frequencies for all classes and the value of chi square. The necessary calculations for expected frequencies , ei`s ( ei= npi`) where pi` is the estimate of pi together with the value of chi square are shown below.

Categories      zi`      P(Z<z)             pi`       Expected         Observed

                                                                    frequency ei    Frequency Oi

1                     -1.39      0.0823    0.0823       10.78                  27

2                   -0.66       0.2546     0.1723         22.57                30

3                    0.07        0.5279      0.2733       35.80                 29

4                   0.808      0.7881       0.2602        34.08                21

5                    1.54        0.937          0.1489        19.51                 24

Next we need to compute the expected frequencies for all classes and the value of chi square. The necessary calculations for expected frequencies , ei`s ( ei= npi`) where pi` is the estimate of pi together with the value of chi square are shown below.

Categories           Expected         Observed       (oi-ei)²/ei

                       frequency ei    Frequency Oi         OBSERVED VALUE

1                            10.78                  27                     24.41

2                          22.57                30                         1.54

3                          35.80                 29                        1.29

4                         34.08                21                           5.02

5                          19.51                 24                         1.033

<u>Total                                              131                   33.293</u>

There are five categories , we have used the sample mean and sample standard deviation , so the number of degrees of freedom is 5-1-2= 2

The critical region is chi square ≥ chi square (0.001)(2) =9.21

<u>CONCLUSION:</u>

Since the calculated value of chi square =9.21  does not fall in the critical region we are unable to reject our null hypothesis and conclude normal distribution provides a good fit for the given frequency distribution.

7 0
4 years ago
The rectangular prism shown has a length of 5 cm, a height of 3 cm, and the width of 4 cm. What is the surface area of this pris
PilotLPTM [1.2K]

Answer:20

Step-by-step explanation:

because you are finding out the base so width times length

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3 years ago
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3. Now combine any like terms. What is the simplified product?
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(x + 1) (x + 2)

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