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SOVA2 [1]
1 year ago
12

Let r= 12 be the reserve rate. Which of the following is the money multiplier?

Mathematics
1 answer:
Studentka2010 [4]1 year ago
3 0

Answer:

B

Step-by-step explanation:

You might be interested in
The solution to the given system of linear equations lies in which quadrant?
Rina8888 [55]
X-3y=6
x+y=2
this is an substitution problem
so first you can do is rewrite the problem by subjection one variable
x=3y+6
then substitute this in the other proble
x+y=2
(3y+6)+y=2
4y+6=2
4y=2-6
4y=-4
y=-1
then substitute the no. in the original equation. 
x=3y+6
x=3(-1)+6
x=-3+6
x=3
now you got the intercepts and you draw the line and check.
it's in the IV quadrant
5 0
1 year ago
Read 2 more answers
the time taken by a student to the university has been shown to be normally distributed with mean of 16 minutes and standard dev
Naya [18.7K]

Answer:

a) 2.84% probability that he is late for his first lecture.

b) 5.112 days

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 16, \sigma = 2.1

a. Find the probability that he is late for his first lecture.

This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{20 - 16}{2.1}

Z = 1.905

Z = 1.905 has a pvalue of 0.9716

1 - 0.9716 = 0.0284

2.84% probability that he is late for his first lecture.

b. Find the number of days per year he is likely to be late for his first lecture.

Each day, 2.84% probability that he is late for his first lecture.

Out of 180

0.0284*180 = 5.112 days

4 0
1 year ago
A member of a student team playing an interactive marketing game received the fol- lowing computer output when studying the rela
nirvana33 [79]

Answer:

p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91

So on this case for the significance level assumed \alpha=0.05 we see that p_v >\alpha so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

Step-by-step explanation:

Let's suppose that we have the following linear model:

y= \beta_o +\beta_1 X

Where Y is the dependent variable and X the independent variable. \beta_0 represent the intercept and \beta_1 the slope.  

In order to estimate the coefficients \beta_0 ,\beta_1 we can use least squares procedure.  

If we are interested in analyze if we have a significant relationship between the dependent and the independent variable we can use the following system of hypothesis:

Null Hypothesis: \beta_1 = 0

Alternative hypothesis: \beta_1 \neq 0

Or in other words we want to check is our slope is significant (X have an effect in the Y variable )

In order to conduct this test we are assuming the following conditions:

a) We have linear relationship between Y and X

b) We have the same probability distribution for the variable Y with the same deviation for each value of the independent variable

c) We assume that the Y values are independent and the distribution of Y is normal  

The significance level assumed on this case is \alpha=0.05

The standard error for the slope is given by this formula:

SE_{\beta_1}=\frac{\sqrt{\frac{\sum (y_i -\hat y_i)^2}{n-2}}}{\sqrt{\sum (X_i -\bar X)^2}}

Th degrees of freedom for a linear regression is given by df=n-2 since we need to estimate the value for the slope and the intercept.  

In order to test the hypothesis the statistic is given by:

t=\frac{\hat \beta_1}{SE_{\beta_1}}

The p value on this case would be given by:

p_v = 2*P(t_{n-2} > |t_{calc}|)= 0.91

So on this case for the significance level assumed \alpha=0.05 we see that p_v >\alpha so then we can conclude that the result is NOT significant. And we don't have enough evidence to reject the null hypothesis.

So on this case is not appropiate say that :"the more we spend on advertising this product, the fewer units we sell" since the slope for this case is not significant.

3 0
1 year ago
Robert is designing a rectangular window based on the amount of wood he has available to make a frame around the window robert w
amm1812

Answer:3

Step-by-step explanation:

3 0
1 year ago
You offer senior citizens a 20% discount on their tune-ups at your gas station. Assuming that an average of 50 senior citizens g
andreyandreev [35.5K]
Percentage of discount given to senior citizens for tune-ups at the gas station = 20%
Average number of senior citizens visiting the gas station per month = 50
Average price of tune-up before discount = $49.95
Total amount that would
have been collected before discount = 49.95 * 50
                                                           = 2497.50 dollars
Total amount of discount given = (20/100) * 2497.50
                                                  = 499.50 dollars
I hope that the procedure is clear enough for you to understand.
6 0
1 year ago
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