What is 5 1/4 divided by 5/8?

Help pls!!!

saul85 [17]

Answer:

48 units

Step-by-step explanation:

8 0

3 years ago

A city's population was 30,200 people in the year 2010 and is growing by 950 people a year. Give a formula for the city's popula

kolbaska11 [484]

the formula should look like this:

p=30200+950t

3 0

3 years ago

Read 2 more answers

4. A toy rocket is fired into the air from the top of a barn. It's height (h) above the ground in

Phantasy [73]

Answer:

A) 25

Step-by-step explanation:

You use x= -b/2a in order to get your maximum so x= -10/ 2(-5) and you get x=1. You plug 1 into the function as h(1)=-5(1)^2 + 10(1)+ 20 which gets you 25

6 0

3 years ago

The mean one-way commute to work in Chowchilla is 7 minutes. The standard deviation is 2.4 minutes, and the population is normal

adell [148]

Answer:

The answer is below

Step-by-step explanation:

Given that:

The mean (μ) one-way commute to work in Chowchilla is 7 minutes. The standard deviation (σ) is 2.4 minutes.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma}$

a) For x < 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

From normal distribution table, P(x < 2) = P(z < -2.08) = 0.0188 = 1.88%

b) For x = 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

For x = 11:

$z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67$

From normal distribution table, P(2 < x < 11) = P(-2.08 < z < 1.67 ) = P(z < 1.67) - P(z < -2.08) = 0.9525 - 0.0188 = 0.9337

c) For x = 11:

$z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67$

From normal distribution table, P(x < 11) = P(z < 1.67) = 0.9525

d) For x = 2:

$z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08$

For x = 5:

$z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83$

From normal distribution table, P(2 < x < 5) = P(-2.08 < z < -0.83 ) = P(z < -0.83) - P(z < -2.08) = 0.2033- 0.0188 = 0.1845

e) For x = 5:

$z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83$

From normal distribution table, P(x < 5) = P(z < -0.83) = 0.2033

8 0

4 years ago

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$