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3 years ago

Seventy and six hundred two thousandths as a decimal number

Mathematics

2 answers:

Alex73 [517]3 years ago

7 0

76.002 i think honestly don’t know but have it a try

Nataliya [291]3 years ago

6 0

Answer:

If I read the question correctly it is 70.062

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Can y’all help me answer these questions

cupoosta [38]

2. $f(x)=\dfrac18\left(\dfrac14\right)^{x-2}$

Domain: $(-\infty,\infty)$ , because any value of $x$ is allowed and gives a number $f(x)$ .

Range: $(0,\infty)$ , because $a^x>0$ for any positive real $a\neq0$ .

y-intercept: This is a point of the form $(0, f(0))$ . So plug in $x=0$ ; we get $f(0)=\frac18\left(\frac14\right)^{-2}=\frac{4^2}8=2$ . So the intercept is (0, 2), or just 2. (Interestingly, you didn't get marked wrong for that...)

Asymptote: This can be deduced from the range; the asymptote is the line $y=0$ .

Increasing interval: Going from left to right, there is no interval on which $f(x)$ is increasing, since 1/4 is between 0 and 1.

Decreasing interval: Same as the domain; $f(x)$ is decreasing over the entire real line.

End behavior: The range tells you $f(x)>0$ , and you know $f(x)$ is decreasing over its entire domain. This means that $f(x)\to+\infty$ as $x\to-\infty$ , and $f(x)\to0$ and $x\to+\infty$ .

3. $f(x)=\left(\dfrac32\right)^{-x}-7$

Domain: Same as (2), $(-\infty, \infty)$ .

Range: We can rewrite $f(x)=\left(\frac23\right)^x-7$ . $\left(\frac23\right)^x>0$ for all $x$ , so $\left(\frac23\right)^x-7>-7$ for all $x$ . Then the range is $(-7,\infty)$ .

y-intercept: We have $f(0)=\left(\frac32\right)^0-7=1-7=-6$ , so the intercept is (0, -6) (or just -6).

Asymptote: $y=-7$

Increasing interval: Not increasing anywhere

Decreasing interval: $(-\infty,\infty)$

End behavior: Similar to (2), but this time $f(x)\to+\infty$ as $x\to-\infty$ and $f(x)\to-7$ as $x\to+\infty$ .

5 0

4 years ago

20 POINTS <br><br> Please only have to answer question 9

KengaRu [80]

Answer:

l = -16/3 m = 80/3

Step-by-step explanation:

see attached for step by step

8 0

2 years ago

I need help with 1-9 please help!!

professor190 [17]

Answer:

Step-by-step explanation:this is so easy

6 0

3 years ago

Absolute value for |27 1/4|

mr_godi [17]

Answer:

27 1/4

Step-by-step explanation:

4 0

3 years ago

The National Transportation Safety Board publishes statistics on the number of automobile crashes that people in various age gro

ra1l [238]

Answer:

a) Null hypothesis: $p\leq 0.12$

Alternative hypothesis: $p > 0.12$

b) $z_{\alpha}=1.64$

c) $z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362$

d) $p_v =P(z>1.362)=0.0866$

e) For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

Step-by-step explanation:

Information given

n=1000 represent the random sample selected

X=134 represent the number of young drivers ages 18 – 24 that had an accident

$\hat p=\frac{134}{1000}=0.134$ estimated proportion of young drivers ages 18 – 24 that had an accident

$p_o=0.12$ is the value that we want to verify

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic

$p_v{/tex} represent the p valuePart aWe want to verify if the population proportion of young drivers, ages 18 – 24, having accidents is greater than 12%: Null hypothesis:[tex]p\leq 0.12$

Alternative hypothesis: $p > 0.12$

The statistic would be given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Part b

For this case since we are conducting a right tailed test we need to find a critical value in the normal standard distribution who accumulates 0.05 of the area in the right and we got:

$z_{\alpha}=1.64$

Part c

For this case the statistic would be given by:

$z=\frac{0.134 -0.12}{\sqrt{\frac{0.12(1-0.12)}{1000}}}=1.362$

Part d

The p value can be calculated with the following probability:

$p_v =P(z>1.362)=0.0866$

Part e

For this case since the statistic is lower than the critical value and the p value higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we don't have information to conclude that the true proportion is higher than 0.12

8 0

3 years ago