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Sauron [17]
3 years ago
12

find the mean and median for the following annual salaries (in thousands) for employees of the Widget Factory 38.5, 31.0, 29.8,

3.4, 40.1, 35.1, 30.8, 41.5, 12.6, 39.7, 28.4, 34.2, 38.6, 35.5, 187.4, 40.6, 39.7, 31.0, 29.8, 42.0 which is the most representative of the average annual salary? why?
Mathematics
1 answer:
Ad libitum [116K]3 years ago
5 0
Mean: add all of them and divide by 20 (for their are 20 numbers)

Median: put them all in order from smallest to greatest and the middle number is your answer ( there are 2 becuz it is an even number)
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Find the GCF of the following literal terms.<br><br> m^7 n^4 p^3 and mn^12 p^5
Dvinal [7]
m^7n^4p^3=mn^4p^3\cdot m^6\\\\mn^{12}p^5=mn^4p^3\cdot n^8p^2\\\\GCF(m^7n^4p^3;\ mn^{12}p^5)=mn^4p^3
6 0
3 years ago
Solve for r please 80=10r.
OLEGan [10]
R = 8 since 8 times 10 = 80
4 0
3 years ago
Read 2 more answers
Please help!! ASAP!!!!
Softa [21]
Tan(theta)= pi

Now, in order for you to find the angle (theta), you would inverse tan both the left and the right to find the value. The inverse of tan is tan^-1.

tan(theta)=pi ---> theta = tan^-1(pi)

Then, you would plug tan^-1(pi) into a calculator, giving you the value 

1.26
8 0
3 years ago
A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic
Ket [755]

Answer:

t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601  

P-value  

First we find the degrees of freedom given by:

df = n-1= 54-1=53

Since is a two-sided test the p value would be:  

p_v =2*P(t_{53}>0.9601)=0.3414  

Step-by-step explanation:

Data given and notation  

\bar X=5.25 represent the sample mean

s=3.98 represent the sample standard deviation

n=54 sample size  

\mu_o =4.73 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean differs from 4.73, the system of hypothesis would be:  

Null hypothesis:\mu =4.73  

Alternative hypothesis:\mu \neq 4.73  

Since we know don't the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601  

P-value  

First we find the degrees of freedom given by:

df = n-1= 54-1=53

Since is a two-sided test the p value would be:  

p_v =2*P(t_{53}>0.9601)=0.3414  

7 0
3 years ago
To graduate this semester, you must pass Statistics 314; and you estimate that you have an 85% chance of passing. You need to pa
Nitella [24]

Answer:

Probability of graduating this semester is 0.7344

Step-by-step explanation:

Given the data in the question;

let A represent passing STAT-314

B represent passing at least in MATH-272 or MATH-444

M1 represent passing in MATH-272

M2 represent passing in MATH-444

C represent  passing GERMAN-32

now

P( A ) = 0.85, P( C ) = 90, P( M1 ) = P( M2 ) = 0.8

P( B ) = P( pass at least one of either MATH-272 or MATH-444 ) = P( pass in MATH-272 but not MATH-444 ) + ( pass in MATH-444 but not in MATH 272) + P( pass both )

P( B ) =  P( M1 ) × ( 1 - P( M2 ) ) + ( 1 - P( M1 ) ) × P( M2 ) + P( M1 ) × P( M2 )

we substitute

⇒ 0.8×0.2 + 0.2×0.8 + 0.8×0.8 = 0.16 + 0.16 + 0.64 = 0.96

∴ the probability of graduating this semester will be;

⇒ P( A ) × P( B ) × P( C )

we substitute

⇒ 0.85 × 0.96 × 90

⇒ 0.7344

Probability of graduating this semester is 0.7344

6 0
3 years ago
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