What is the true solution to the equation below?<br> 2 In eln2x

It has been found that 85.6% of all enrolled college

NNADVOKAT [17]

Answer:

it has been found that 85.6% of all enrolled college and university students in the U.S. are undergraduates. A random sample of 500 enrolled college students in a particular state revealed that 420 of them were undergraduates. Is there sufficient evidence to conclude that the population differs from the national percentages? Use a = 0.05

6 0

3 years ago

Whats 981.992 rounded to the nearest whole number

gladu [14]

982! Hope I helped you out.

3 0

3 years ago

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Write the equation of the line that passes through the points (2 , -1) and has a slope of - 3

svetlana [45]

Answer:

y=-3x+5

Step-by-step explanation:

y=mx+b

m=slope or gradient (=-3)

y=-3x+b

Add your points

(2, -1)

-1=(-3 x 2)+b

-1= -6+b

-1+6=b

b=5

y=-3x+5

5 0

3 years ago

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find the number of ways the arrangements can be made of the letters of the word'WONDERFUL' such that the letter R is always next

zalisa [80]

Answer:

The number of ways the arrangements can be made of the letters of the word'WONDERFUL' such that the letter R is always next to E is 10,080 ways

Step-by-step explanation:

We need to find the number of ways the arrangements can be made of the letters of the word'WONDERFUL' such that the letter R is always next to E.

There are 9 letters in the word WONDERFUL

There is a condition that letter R is always next to E.

So, We have two letters fixed WONDFUL (ER)

We will apply Permutations to find ways of arrangements.

The 7 letters (WONDFUL) can be arranged in ways : ⁷P₇ = 7! = 5040 ways

The 2 letters (ER) can be arranged in ways: ²P₂ =2! = 2 ways

The number of ways 'WONDERFUL' can be arranged is: (5040*2) = 10,080 ways

So, the number of ways the arrangements can be made of the letters of the word'WONDERFUL' such that the letter R is always next to E is 10,080 ways

8 0

3 years ago

ASK YOUR TEACHER An article reported that for a sample of 42 kitchens with gas cooking appliances monitored during a one-week pe

xxMikexx [17]

Answer:

$654.16-2.02\frac{165.23}{\sqrt{42}}=602.66$

$654.16+2.02\frac{165.23}{\sqrt{42}}=705.66$

So on this case the 95% confidence interval would be given by (602.66;705.66)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=654.16$ represent the sample mean

$\mu$ population mean (variable of interest)

s=165.23 represent the sample standard deviation

n=42 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=42-1=41$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,41)".And we see that $t_{\alpha/2}=2.02$

Now we have everything in order to replace into formula (1):

$654.16-2.02\frac{165.23}{\sqrt{42}}=602.66$

$654.16+2.02\frac{165.23}{\sqrt{42}}=705.66$

So on this case the 95% confidence interval would be given by (602.66;705.66)

7 0

4 years ago

What is the true solution to the equation below? 2 In eln2x - In el 10x = In 30