Choose all the expressions that have 3/4 as a product.

A linear model projects that the number of bachelor's degrees awarded in the United State will increase by 19,500 each year. In

vichka [17]

Let the number of years after 2001 be x, then
1280000 + 19500x = 1700000
19500x = 1700000 - 1280000 = 420,000
x = 420000/19500 = 22 years.
x = 22 years.

6 0

3 years ago

The total cost to attend the university Daniel wants to attend is going to be $12,000 for the first year.

Zielflug [23.3K]

Answer:

$395.83

Step-by-step explanation:

to solve, we first need to subtract what we already have, first the scholarship, wich is a set amount being taken from the original amount

12'000 - 2'500 = 9'500

now we have the second subtracting factor, but this one isn't set in stone and defined, it is half of the amount his parents will pay, so what we can do is divide what we have by two, wich will give us 2 halves

9'500 / 2 = 4'750

now all we have to do is divide again but this time for each month that Daniel needs to save up in, in this case 12

4'750 / 12 = 395.83 (note1)

and there we have it, that is the minimum amount Daniel would save each month

note1: (3 goes on for infinity, usualy this is represented by a line above the repeating number, this is the case of a repeating decimal)

5 0

3 years ago

Using traditional methods, it takes 10.1 hours to receive a basic driving license. A new license training method using Computer

Law Incorporation [45]

Answer:

We need to conduct a hypothesis in order to check if the technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

$p_v =2*P(t_{(11)}>0.990)=0.343$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=10.5$ represent the sample mean

$s=1.4$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =10.1$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t 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 technique performs differently than the traditional method, the system of hypothesis would be:

Null hypothesis: $\mu =10.1$

Alternative hypothesis: $\mu \neq 10.1$

If we analyze the size for the sample is < 30 and we don't know the population deviation so 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:

$t=\frac{10.5-10.1}{\frac{1.4}{\sqrt{12}}}=0.990$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=12-1=11$

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

$p_v =2*P(t_{(11)}>0.990)=0.343$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can't conclude that the technique performs differently than the traditional method at 10% of signficance.

7 0

3 years ago

Listed below are the weights in pounds of 11 players randomly selected from the roster of the Seattle Sea-hawks when they won Su

weeeeeb [17]

Answer:

No. These measures show a thinner team of NFL players according to the mean, variance, standard deviation, and quartiles.

Step-by-step explanation:

1) The measures of variation, namely The Range, Variance, Quartiles, Interquartiles, Sum of Squares, etc. shows us how the data are dispersed.

The Range Δ is calculated:

$\Delta =305-189=116$ Maximum value - Minimum value for weight

Mean:

$\bar{x}=\frac{189+254+235+225+190+305+202+190+252+305}{11}\approx 231.09 \:pounds$

Variance:

$s^{2}=\frac{\sum_{i=1}^{n} (x_i-\bar{x})^{2}}{n-1}\Rightarrow s=1923.69\\$

The Standard Deviation of the sample

$s=\sqrt{s^{^{2}}}\Rightarrow s\approx 43.86\\$

2) Since there is no preceding exercise, the comparison was made to a recent study in which a NFL player average weight is about 245 pounds (average),

Since 25% of this list are player whose weight is 192.5 lbs and 50% (2nd Quartile) =225 lbs , finally only at the 3rd Quartile we have players above the regular NFL average with 253. This, added with the other data, allow us to say that this list is not a typical of all NFL players.

4 0

3 years ago

There are 250 students taking band, chorus, or both. If there are 180 students taking band and 60 students in both band and chor

katrin2010 [14]

I think your answer here would be 70.

8 0

4 years ago