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Bingel [31]
2 years ago
12

Please answer these 2 questions

Mathematics
2 answers:
podryga [215]2 years ago
8 0

Question 5: /  Pregunta 5:

<u>b. x² + 3</u>

f(x) = \sqrt{x-3}

\/

y = \sqrt{x-3}

\/

x = \sqrt{y-3}

\/

x² = \sqrt{y-3}^{2}

\/

x² = y - 3

\/

x² + 3 = y

\/

x² + 3 = f(x)

Question 6: /  Pregunta 6:

<u>b. Yes, but not one to one</u>

        Two inputs have an output of 8 so it is not one to one, but no input has two different outputs so it is a function / Dos entradas tienen una salida de 8, por lo que no es una a una, pero ninguna entrada tiene dos salidas diferentes, por lo que es una función.

dolphi86 [110]2 years ago
5 0

Question 5: / Pregunta 5:

b. x² + 3

f(x) = \sqrt{x-3}

x−3

\/

y = \sqrt{x-3}

x−3

\/

x = \sqrt{y-3}

y−3

\/

x² = \sqrt{y-3}^{2}

y−3

2

\/

x² = y - 3

\/

x² + 3 = y

\/

x² + 3 = f(x)

Question 6: / Pregunta 6:

b. Yes, but not one to one

Two inputs have an output of 8 so it is not one to one, but no input has two different outputs so it is a function / Dos entradas tienen una salida de 8, por lo que no es una a una, pero ninguna entrada tiene dos salidas diferentes, por lo que es una función.

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129.50

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PLEASE HELP THANK YOU
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Answer:

change in the total cost for each book printed = $15

cost to get started (before books are bought) = $1200

Step-by-step explanation:

IF the eqn is y = 15x +1200

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x  represent the number of copies of the book printed


then cost to get started (before books are bought) can be found by substitute x=0 into the eqn

y = 1200 + 15*0 = $1200


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7 0
3 years ago
A researcher believes that there is a difference in how cautious male stray cats and female stray cats are when approaching peop
Anika [276]

Answer:

The null hypothesis is that there is no difference in the mean number of male and female cats

H₀; μ₂ - μ₁ = 0

Step-by-step explanation:

The given parameters are;

The given percentage of male stray cat population = 50%

The given percentage of female stray cat population = 50%

The number of areas the researcher visits, n = 15

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The kind of test to be performed = Sign test

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5 0
2 years ago
Suppose that the probabilities that an answer can be found on Google is .95, on Answers is .92, and on both Web sites is .874. A
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Answer:

The correct option is

a) Yes, because (0.95)(0.92) = 0.874

Step-by-step explanation:

An event is independent if the following condition is satisfied

P(A\bigcap B)= P(A)P(B)

Therefore, since the the probability that the answer can be found in Google = 0.95 and the probability that the answer can be found in Answers = 0.92

The probability that the answer can be found on both websites,  

P(A\bigcap B) = 0.874

We check, P(A) × P(B) = 0.95×0.92 = 0.874 = P(A\bigcap B) = 0.874

Therefore, the possibilities of finding the event on the two websites is independent.

5 0
2 years ago
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Alja [10]

Answer:

Down below.

Step-by-step explanation:

Use Pythagorean’s Theron

9 + 16 = c squared

25 = c squared

c = 5

6 0
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