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3 years ago

What is the range of y=log^2(x-6)

Mathematics

2 answers:

Alborosie3 years ago

8 0

Answer:

Ans: All real numbers

Step-by-step explanation:

Marysya12 [62]3 years ago

6 0

Answer:

$y\geq 0$ or $[0, \infty)$

Step-by-step explanation:

By definition the function

$y = log (x)$ has a domain of $x> 0$ .

Since the function is not defined for the values of x negative or equal to zero.

The range of this function is all the real numbers.

Since $y$ when $0$ and $y\geq0$ when $1\leq x$

In this case we have the function $y = log ^ 2 (x-6)$

Therefore since the log function is squared then its range is now $y\geq 0$

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The number of goals scored by a soccer team varies directly as the number of shots on goal. If the Lakeside Bears scored 4 goals

enot [183]

Since it varies directly, divide the number of shots by the number of goals:

48 shots / 4 goals = 12

This tells you they score a goal every 12 shots.

Multiply shots needed by number of goals:

12 x 3 = 36

They need 36 more shots to score 3 more goals.

6 0

2 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

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

= $P(Z\leq -2.3)$

= .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

Someone please help me

Lilit [14]

<h3>Answer:</h3>

C. (9x -1)(x +4) = 9x² +35x -4
B. 480
A. P(t) = 4(1.019)^t

Step-by-step explanation:

1. See the attachment for the filled-in diagram. Adding the contents of the figure gives the sum at the bottom, matching selection C.

2. If we let "d" represent the length of the second volyage, then the total length of the two voyages is ...

... (d+43) + d = 1003

... 2d = 960 . . . . . . . subtract 43

... d = 480 . . . . . . . . divide by 2

The second voyage lasted 480 days.

3. 1.9% - 1.9/100 = 0.019. Adding this fraction to the original means the original is multiplied by 1 +0.019 = 1.019. Doing this multiplication each year for t years means the multiplier is (1.019)^t.

Since the starting value (in 1975) is 4 (billion), the population t years after that is ...

... P(t) = 4(1.019)^t