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3 years ago

14

So can someone show there work in this one?

1 answer:

mihalych1998 [28]3 years ago

4 0

just separate the ramge and domain

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Julie knows that the adult population gets, on average, eight hours of sleep each night. A hypothesis test can help her see if c

Mamont248 [21]

Answer:

-3.64

Step-by-step explanation:

6 0

3 years ago

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

5 0

3 years ago

Read 2 more answers

Explain how to model the division of –20 by –5 on a number line.

kenny6666 [7]

Star at -20 and go back by -5 on the number line

3 0

3 years ago

.A variety of stores offer loyalty programs. Participating shoppers swipe a bar-coded tag at the register when checking out and

Leni [432]

Answer:

a) Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

b) $t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

Step-by-step explanation:

Information given

$\bar X=130$ represent the sample mean for the amount spent each shopper

$s=40$ represent the sample standard deviation

$n=80$ sample size

$\mu_o =120$ represent the value to verify

t would represent the statistic

$p_v$ represent the p value f

Part a

We want to verify if the shoppers participating in the loyalty program spent more on average than typical shoppers, the system of hypothesis would be:

Null hypothesis: $\mu \leq 120$

Alternative hypothesis: $\mu > 120$

The statistic for this case would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{130-120}{\frac{40}{\sqrt{80}}}=2.236$

The degrees of freedom are given by:

$df = n-1 = 80-1=79$

The p value for this case taking in count the alternative hypothesis would be:

$p_v =P(t_{79}>2.236)=0.0141$

5 0

3 years ago

HELP ON FIRST 3 PLS

rodikova [14]

Answer:

THE answer are

A' ( -5,-2 )
G' (-8,1 )
E' (4 -9 )

HOPE IT HELP YOU.

4 0

2 years ago