Next 4 multiples of 1/6

Jesse worked forty hours this past week assuming he makes $12 an hour how much was withheld from his paycheck by his employer fo

Alecsey [184]

He made 12*40=480 dollars in total. 0.0765*480=36.72 dollars were withheld.

6 0

4 years ago

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What is the length of BD Round to one decimal place. Thanks!

Assoli18 [71]

Answer:

2.7

Step-by-step explanation:

ratios help

2.5 : 5.8 :: x : 6.2

2.5/5.8 = x/6.2

solve for x :

x = approx. 2.7

8 0

3 years ago

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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

Calculate the average rate of change of f(x) = 1 x - x2 - 2 for 3 ≤ x ≤ 6. A) -163 18 B) -18 163 C) 163 18 D) 18 163

larisa86 [58]

To solve this we are going to use the average rate of change formula: $A(x)= \frac{f(b)-f(a)}{b-a}$
where
$A(x)$ is the average rate of change of the function
$f(a)$ is the position function evaluated at $a$
$f(b)$ is the position function evaluated at $b$
$a$ is the first point in the interval
$b$ is the second point in the interval

We can infer for our problem that the first point is 3 and the second point is 6, so $a=3$ and $b=6$ . Lets replace those values in our formula:
$A(x)= \frac{f(b)-f(a)}{b-a}$
$A(x)= \frac{f(6)-f(3)}{6-3}$
$A(x)= \frac{6-6^2-2-(3-3^2-2)}{3}$
$A(x)= \frac{-32-(-8)}{3}$
$A(x)= \frac{-32+8}{3}$
$A(x)= \frac{-24}{3}$
$A(x)=-8$

We can conclude that the average rate of change of the function f(x) = 1 x - x2 - 2 for <span>3 ≤ x ≤ 6 is -8</span>

3 0

3 years ago

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A simple random sample of 60 items resulted in a sample mean of 65. The population standard deviation is 14. a. Compute the 95%

bazaltina [42]

Answer:

The 95% confidence interval for the population mean is between 61.5 and 68.5.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So: