Please help I will mark 5 stars and worth 30 pts

Which statement is true regarding the graphed functions?

Vladimir79 [104]

let's analyze each case to determine the solution

case 1) f(0) = 2 and g(–2) = 0

For x=0-----> find the value of f(0) in the graph-----> f(0)=4

For x=-2-----> find the value of g(-2) in the graph-----> g(-2)=0

therefore

the statement of the case 1) is false

case 2) f(0) = 4 and g(–2) = 4

For x=0-----> find the value of f(0) in the graph-----> f(0)=4

For x=-2-----> find the value of g(-2) in the graph-----> g(-2)=0

therefore

the statement of the case 2) is false

case 3) f(2) = 0 and g(–2) = 0

For x=2-----> find the value of f(2) in the graph-----> f(2)=0

For x=-2-----> find the value of g(-2) in the graph-----> g(-2)=0

therefore

the statement of the case 3) is true

case 4) f(–2) = 0 and g(–2) = 0

For x=-2-----> find the value of f(-2) in the graph-----> f(-2) is greater than 12

For x=-2-----> find the value of g(-2) in the graph-----> g(-2)=0

therefore

the statement of the case 4) is false

therefore

the answer is

f(2) = 0 and g(–2) = 0-------> this statement is true

6 0

4 years ago

Read 2 more answers

21.60 divided by 1.35 is what?

klasskru [66]

Answer:

16

Step-by-step explanation:

21.60 divided by 1.35=16

8 0

3 years ago

Read 2 more answers

NarStor, a computer disk drive manufacturer, claims that the median time until failure for their hard drives is more than 14,400

Ne4ueva [31]

Answer:

The test statistics is $t = -1.727$

Step-by-step explanation:

From the question we are told that

The data given is

330 620 1870 2410 4620 6396 7822 81028309 12882 14419 16092 18384 20916 23812 25814

The population mean is $\mu = 14400$

The sample size is n = 16

The null hypothesis is $\mu \le 14400$

The alternative hypothesis is $H_a : \mu > 14400$

The sample mean is mathematically evaluated as

$\= x = \frac{\sum x_i}{n}$

So

$\= x = \frac{330+ 620+ 1870 +2410+ 4620+ 6396+ 7822+ 8102+8309+ 12882+ 14419+ 16092+ 18384 +20916+ 23812+ 25814 }{16}$

=> $\= x = 10799.9$

The standard deviation is mathematically represented as

$\sigma =\sqrt{\frac{ \sum (x_i - \=x)^2}{n}}$

So

$\sigma =\sqrt{\frac{(330- 10799.9)^2 + (620- 10799.9)^2+ (1870- 10799.9)^2 +(2410- 10799.9)^2 + (4620- 10799.9)^2 +(6396- 10799.9)^2 +(7822- 10799.9)^2 }{16}} \ ..$

$..\sqrt{ \frac{(8102 - 10799.9)^2 +(8309 - 10799.9)^2 + (12882 - 10799.9)^2 + (14419 - 10799.9)^2 + (16092 - 10799.9)^2 + (18384 - 10799.9)^2 +(20916 - 10799.9)^2 }{16}} \ ...$