Order from greatest to least. 5.249; 5.43; 5.340; 5.209

Suppose the average yearly salary of an individual whose final degree is a master's is $46 thousand less than twice that of an

seraphim [82]

Answer:

The average sallary of a Master's is 60 thousand and of a Bachellor's is 53 thousand.

Step-by-step explanation:

In order to solve this problem we first need to attribute variables to the unkown quantities. We will call the average salary of Master's "x" and the average salary of a Bachellor's "y". The first information the problem gives us is:

x = 2*y - 46

The second one is:

x + y = 113

We now have two equations and two variables, so we can solve the system. To do that we will use the value for x from the first equation on the second one. We have:

2*y - 46 + y = 113

3*y = 113 + 46

3*y = 159

y = 159/3 = 53

x = 2*(53) - 46 = 60

The average sallary of a Master's is 60 thousand and of a Bachellor's is 53 thousand.

3 0

4 years ago

What is the number of possible outcomes if three quarters are tossed and the total numbers of heads and tails are counted? A) 2

Gre4nikov [31]

Answer:

D). 8.

Step-by-step explanation:

Each quarter will come up heads or tails - that is 2 possibilities.

So for 3 tosses the number of outcomes is 2*2*2 = 8.

3 0

3 years ago

Read 2 more answers

Help please!!!

Tanya [424]

Sin C = AB / AC
We have AC. We need to calculate AB

ABC is a right triangle. Therefore, AC² = AB²+BC²
82² = AB ² + 80² => AB² = 82² - 80²
AB = 18

Sin C = 18/82 = 9/41

Good luck

4 0

3 years ago

Read 2 more answers

The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard de

Sergio [31]

To solve this problem, we use the formula:

z = (x – u) / s

where z is the z score value which can be obtained from the tables, x is the sample value, u is the mean = 6.3 min, and s is the std dev = 2.2 min

at P value = 0.90, the z = 1.28, finding for x:

x = z s + u

x = 1.28 * 2.2 + 6.3

x = 9.116

at P value = 1.0, the z = 3.49, finding for x:

x = z s + u

x = 3.49 * 2.2 + 6.3

x = 13.978 ~ 14

Therefore the longest 10% calls last about 9.1 minutes to 14 minutes

3 0

4 years ago

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

laila [671]

Answer:

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

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".

$\bar X=17$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.9 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

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:

$df=n-1=10-1=9$

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(0.025,9)".And we see that $t_{\alpha/2}=2.262$

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

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

4 0

3 years ago