PLEASE HELP SOON!! AS POSSIBLE

Last year 49 people adopted a manatee through fia’s foundation. This year 15% more people adopted a manatee. How many people ado

borishaifa [10]

The equation is 49 times .15 which equals 7.35 then you add that to 49 which equals 56.35. Hope that helps and sorry if its wrong

6 0

3 years ago

Problem: Try this one on your own.

Anna [14]

Answer:

iiiii

Step-by-step explanation:

8 0

2 years ago

The 2008 Workplace Productivity Survey, commissioned by LexisNexis and prepared by WorldOne Research, included the question, "Ho

vitfil [10]

Answer:

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

Step-by-step explanation:

Let X be the number of hours a legal professional works on a typical workday. Imagine that X is normally distributed with a known standard deviation of 12.6.

The population standard deviation is

$\sigma = 12.6 \: hours$

A sample of 250 legal professionals was surveyed, and the sample's mean response was 9 hours.

The sample size is

$n = 250$

The sample mean is

$\bar{x} = 9 \: hours$

Since the sample size is quite large then according to the central limit theorem, the sample mean is approximately normally distributed.

The population mean would be the same as the sample mean that is

$\mu = \bar{x} = 9 \: hours$

The sample standard deviation would be

$s = {\frac{\sigma}{\sqrt{n} }$

Where is the population standard deviation and n is the sample size.

$s = {\frac{12.6}{\sqrt{250} }$

$s = 0.7969 \: hours$

Therefore, the sampling distribution of $\bar{x}$ is normal with a mean equal to 9 hours and a standard deviation of 0.7969 hours.

The population mean confidence interval is given by

$\text {confidence interval} = \mu \pm MoE\\\\$

Where the margin of error is given by

$$ MoE = t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $ \\\\$

Where n is the sampling size, s is the sample standard deviation and is the t-score corresponding to a 95% confidence level.

The t-score corresponding to a 95% confidence level is

Significance level = α = 1 - 0.95 = 0.05/2 = 0.025

Degree of freedom = n - 1 = 250 - 1 = 249

From the t-table at α = 0.025 and DoF = 249

t-score = 1.9695

$MoE = t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\\\MoE = 1.9695\cdot \frac{12.6}{\sqrt{250} } \\\\MoE = 1.9695\cdot 0.7969\\\\MoE = 1.569\\\\$

So the required 95% confidence interval is

$\text {confidence interval} = \mu \pm MoE\\\\\text {confidence interval} = 9 \pm 1.569\\\\\text {LCI } = 9 - 1.569 = 7.431\\\\\text {UCI } = 9 + 1.569 = 10.569$

The 95% interval estimate of the population mean $\mu$ is

LCL = 7.431 hours to UCL = 10.569 hours

8 0

3 years ago

Which of the following are solutions to the inequality below? Select all that apply.

Naddik [55]

Answer:

p=4

Step-by-step explanation:

because 5 is less than or equal to four!

another number grater than five no less than

so the answer is p=4

7 0

2 years ago

The cost of petrol rises by 2 cents a liter. Last week a man bought 20 liters at the old price. This week he bought 10 liters at

In-s [12.5K]

Answer: the old price for 1 liter is $0.3

Step-by-step explanation:

Let x represent the old price for 1 liter of petrol.

Last week a man bought 20 liters at the old price. This means that the total amount that the man paid for 20 liters is 20x.

The cost of petrol rises by 2 cents a liter. This means that the new price per liter is x + 2/100 = x + 0.02

This week he bought 10 liters at the new price. This means that the total amount that the man paid for 10 liters is

10(x + 0.02)

= 10x + 0.2

Altogether, the petrol costs $9.20. This means that

20x + 10x + 0.2 = 9.2

30x + 0.2 = 9.2

30x = 9.2 - 0.2

30x = 9

x = 9/30

x = 0.3

3 0

3 years ago