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arlik [135]
3 years ago
10

Four members from a ​"55"person committee are to be selected randomly to serve as​ chairperson, vice-chairperson,​ secretary, an

d treasurer. The first person selected is the​ chairperson; the​ second, the​ vice-chairperson; the​ third, the​ secretary; and the​ fourth, the treasurer. How many different leadership structures are​ possible?
Mathematics
1 answer:
liq [111]3 years ago
3 0

Answer:

8,185,320 different leadership structures

Step-by-step explanation:

Since the order at which the members of the committee are chosen matters, this is a permutation of 4 out 55 people and it is given by:

n=\frac{55!}{(55-4)!}=55*54*53*52 \\\\n=8,185,320

8,185,320 different leadership structures are​ possible.

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Most US adults have social ties with a large number of people, including friends, family, co-workers, and other acquaintances. I
xxTIMURxx [149]

Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. 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 mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but 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{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

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

df=n-1=1700-1=1699  

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

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

3 0
3 years ago
What ratios are equivalent to 5/9
NNADVOKAT [17]

Answer:

10/18 and 20/36

Step-by-step explanation:

5 0
2 years ago
Teresa drove on monday tuesday and Wednesday. On monday her driving distance was 120 miles. The ratio of Wednesdays distance is
SSSSS [86.1K]

Answer:

Teresa drive <u>90 miles</u> on Tuesday.

Step-by-step explanation:

Given:

Teresa drove on Monday, Tuesday and Wednesday.

On monday her driving distance was 120 miles.

The ratio of Wednesdays distance is 3/5.

The ratio of Wednesdays distance to Tuesday's distance is 5/4.

Now, to find the miles Teresa drive on Tuesday:

Teresa driving distance on Monday was = 120 miles.

Her driving distance on Wednesday is = \frac{3}{5}\ of\ 120.

=\frac{3}{5} \times 120\\\\=0.6\times 120\\\\=72\ miles.

Now, her driving distance on Tuesday is:

\frac{5}{4} \ of\ 72\\\\=\frac{5}{4} \times 72\\\\=1.25\times 72\\\\=90\ miles.

Therefore, Teresa drive 90 miles on Tuesday.

7 0
3 years ago
the population of a town increases at the rate of 1% each year today the towns population is 8500 what will the population be in
romanna [79]

Answer:

8934 (8933.58542585)

Step-by-step explanation:

8,500 (population currently) * 1.01 (1 percent) ^ 5 (years) = 8933.58542585 ≅ 8934 people.

4 0
3 years ago
Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required
xenn [34]

Answer:

(a) 8.15

(b) 12.92

Step-by-step explanation:

Given: P = $3000, r = 0.085

                A = Pe^{rt}

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

               A = 2 × $3000

               A = $6000

               6000 = 3000e^{0.085t}

               \frac{6000}{3000} = e^{0.085t}

               2 = e^{0.085t}

Take log_{e} of both sides

               log_{e}2 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln2 = 0.085t

               t = \frac{ln2}{0.085}

               t = \frac{0.693}{0.085}

               t = 8.15

(b) For the amount to double, A = 3 × P

               A = 3 × $3000

               A = $9000

               9000 = 3000e^{0.085t}

               \frac{9000}{3000} = e^{0.085t}

               3 = e^{0.085t}

Take log_{e} of both sides

               log_{e}3 = log_{e}e^{0.085t}

But log_{e}e = 1

            ∴ ln3 = 0.085t

               t = \frac{ln3}{0.085}

               t = \frac{1.0986}{0.085}

               t = 12.92

5 0
3 years ago
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