Pls answer this I need it quick

What are the differences between 0.08 and 0.8

Anon25 [30]

0.8 is greater than 0.08, the reason for that is that 0.8 is equal to 8/10=80/100 and 0.08 is equal to 8/100. As a percentage, 0.8 would be 80% and 0.08 would be equal to 8%.

5 0

3 years ago

The mean consumption of water per household in a city was 1425 cubic feet per month. Due to a water shortage because of a drough

liberstina [14]

Answer:

a) $t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

b) For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

Step-by-step explanation:

Information given

$\bar X=1175$ represent the sample mean for the cubic feets of households

$\sigma=250$ represent the population standard deviation

$n=100$ sample size

$\mu_o =1425$ represent the value to verify

$\alpha=0.025$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test that the mean consumption of water per household has decreased due to the campaign by the city council, the system of hypothesis would be:

Null hypothesis: $\mu \geq 1425$

Alternative hypothesis: $\mu < 1425$

Since we don't know the deviation the statistic is given by:

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

The statistic is given by:

$t=\frac{1175-1425}{\frac{250}{\sqrt{100}}}=-10$

The degrees of freedom are given by:

$df=n-1=100-1=99$

The p value for this case would be given by:

$p_v =P(t_{99}$

Since the p value is significantly lower than he significance level given we have enough evidence to reject the null hypothesis and we can conclude that the true mean is lower than 1425

Part b

For this case we need to find a critical value in the t distribution with 99 degrees of freedom who accumulates 0.025 of the area in the right tail and we got:

$t_{\alpha/2}=1.984$

Since the calculated value is higher than the critical value we can reject the null hypothesis at the significance level provided and we can say that the true mean is higher than 1425

6 0

3 years ago

.0417 DIVIDED 100 PLZ HELP (⊙_⊙)

nikklg [1K]

Answer: 0.000417

Hope this helps can u give me brainliest

Step-by-step explanation:

7 0

3 years ago

The number of calculators and backpacks that were turned into the school lost and found have been recorded for 10 days in the sc

34kurt

You need to show the scatter plot

8 0

2 years ago

Can someone please help me on this??

TEA [102]

X= base

Area= 1/2(base)(height)
24=1/2(b)(4)
24=(4*1*b)/2
24=4b/2
Divide 2 into 4
24=2b
Divide both sides by 2
12=b

base= x= 12 units

Hope this helps! :)

4 0

3 years ago