47% of 118% the estimate of it

Determine the intercepts of the line.<br> 71 - 5 = 4y - 6

maw [93]

Answer:

y = 18

Step-by-step explanation:

71 - 5 = 4y - 6

66 = 4y - 6

72 = 4y

18 = y

P.S. I am not 100% sure if this is correct if it isn't I am greatly sorry

8 0

3 years ago

Week 2 4.0A.A.3 V 27 X 34

zalisa [80]

Answer:

is there a question for this problem if there is try your best

8 0

3 years ago

The population of a certain city was 25000 in 2010 and decreased by 4% each year afterword. The population can be modeled by an

meriva

a= initial value, 25000. b = rate of decay, (1-.04) or (.96)
a represents the initial value in exponential functions and b represents the rate of growth or decay. for decay subtract the 4% from 1 or 1-.04.

3 0

3 years ago

PLEASE HELP!!!

Nuetrik [128]

The correct aswers is A

5 0

3 years ago

The magic dragon cigarette company claims that their cigarettes contain an average of only 10 mg of tar. a random sample of 100

9966 [12]

Answer:

$t=\frac{11.5-10}{\frac{4.5}{\sqrt{100}}}=3.333$

Critical value

The significance is 5% so then $\alpha=0.05$ and $\alpha/2=0.025$ then the critical value for this case would be $z_{\alpha/2}= 1.64$ . Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true mean is higher than 10 mg

P value

The p value would be given

$p_v =P(z>3.333)=0.000434$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 10 mg

Step-by-step explanation:

Information given

$\bar X=11.5$ represent the sample mean

$\sigma=4.5$ represent the population standard deviation

$n=100$ sample size

$\mu_o =10$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean is higher than 10mg, the system of hypothesis would be:

Null hypothesis: $\mu \leq 10$

Alternative hypothesis: $\mu > 10$

the statistic for this case would be given by:

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

Replacing we got:

$t=\frac{11.5-10}{\frac{4.5}{\sqrt{100}}}=3.333$

Critical value

The significance is 5% so then $\alpha=0.05$ and $\alpha/2=0.025$ then the critical value for this case would be $z_{\alpha/2}= 1.64$ . Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis and we can conclude that the true mean is higher than 10 mg

P value

The p value would be given

$p_v =P(z>3.333)=0.000434$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 10 mg

6 0

3 years ago