Question 9:

This is only the first question out of many

klemol [59]

3, 5, 7
wanna know why, or do you not care?

4 0

4 years ago

Express… As a limit of a right Reimann sum

kobusy [5.1K]

answer

the answer is 36

3 0

2 years ago

3. Find f[h(a+4)], given the functions below. *

mote1985 [20]

Answer:

C 5a^2 +70a +240

Step-by-step explanation:

We have given these following functions:

$f(x) = 5x, g(x) = -2x + 1, h(x) = x^2 + 6x + 8$

h(a+4)

This function is:

$h(a+4) = (a+4)^2 + 6(a + 4) + 8$

$h(a+4) = a^2 + 8a + 16 + 6a + 24 + 8$

$h(a+4) = a^2 + 14a + 48$

f[h(a+4)]

$f(x) = 5x$

Thus

$f(h(a+4)) = f(a^2+14a+48) = 5(a^2 + 14a + 48) = 5a^2 + 70a + 240$

The correct answer is given by option C.

8 0

3 years ago

On the last day of school the fifth grade class has a picnic at the city park. Exactly 182 students and 274 adults are signed up

mamaluj [8]

Answer:

38 Tables

Step-by-step explanation:

First, we need to add the number of students with the number of adults to get the total number of people attending the picnic.

182 + 274 = 456

Next, we divide that number by how many people fit at each table to find how many tables we need.

456 people ÷ 12 people at each table = 38 tables.

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