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3 years ago

Is before 75 and after 60 a decrease in %?

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

7 0

Answer:

20%

Step-by-step explanation:

Percent decrease = [(original value - new value)/original value * 100]

Percent decrease = [(75 - 60)/75 * 100]

Percent decrease = [15/75 * 100] = 0.2*100= 20

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Let Xi,X2,X3,... be i.i.d. Bernoulli trials with success probability p and Sk=X1+.....+Xk. Let m< n.

Kay [80]

Answer:

Detailed step wise solution is given below:

Step-by-step explanation:

If X_i,i=1,2,3,... are Bernoulli random variables, then its PMF is

P\left (X_i =1 \right )=p, P\left (X_i =0 \right )=1-p,i=1,2,3,...

Define S_k=X_1+X_2+...+X_k . When S_n=k,0\leqslant k\leqslant n. Then k out of n random variables equals to 1. There are \binom{n}{k} possible combinations of k 1's and n-k 0's. So we have

P\left ( S_n=k \right )=\binom{n}{k}p^k\left ( 1-p \right )^{n-k},k=0,1,2,...,n . That is S_n has Binomial distribution.

a)The joint probability mass function of random vector \left ( X_1,X_2,...,X_m \right ) given S_n=X_1+X_2+...+X_n=k defined as \left (n\geqslant m \right )

P\left ( X_1=a_1,X_2=a_2,...,X_m=a_m|S_n=k \right ) can be calculated as below.

P\left ( S_m=l,S_n=k \right )=\binom{m}{l}p^l\left ( 1-p \right )^{m-l}\binom{n-m}{k-l}p^{k-l}\left ( 1-p \right )^{n-m-k+l}\\ P\left ( S_m=l,S_n=k \right )=\binom{m}{l}\binom{n-m}{k-l}p^k\left ( 1-p \right )^{n-k};l=0,1,2,..,m;k=l,..,n

The conditional distribution,

P\left ( S_m=l|S_n=k \right )=\frac{P\left ( S_m=l,S_n=k \right )}{P\left ( S_n=k \right )}\\ P\left ( S_m=l|S_n=k \right )=\frac{\binom{m}{l}\binom{n-m}{k-l}p^k\left ( 1-p \right )^{n-k}}{\binom{n}{k}p^k\left ( 1-p \right )^{n-k}}\\ {\color{Blue} P\left ( S_m=l|S_n=k \right )=\frac{\binom{m}{l}\binom{n-m}{k-l}}{\binom{n}{k}};l=0,1,2,..,m;k=l,..,n}

This distribution is Hyper geometric distribution. We have to get l successes in first m trials and k-l successes in the next n-m trials. The total ways of happening this is \binom{n}{k} . Hence Hyper geometric.

b) The conditional expectation is

E\left ( S_m=l|S_n=k \right )=\sum_{l=0}^{m}lP\left ( S_m=l|S_n=k \right )\\ E\left ( S_m=l|S_n=k \right )=\sum_{l=0}^{m}l\times \frac{\binom{m}{l}\binom{n-m}{k-l}}{\binom{n}{k}}\\

Use the formula for expectation of hyper geometric distribution, {\color{Blue} E\left ( S_m=l|S_n=k \right )=\frac{k m}{n}}

7 0

4 years ago

Drag the tiles to the correct boxes to complete the pairs.

loris [4]

Answer:

Match each situation to the simulation that represents it.

Althea has to select 1 student

as the class representative

from 13 students.

Fiona is deciding which of three

forests to research for a

geography project: evergreen,

deciduous, or coniferous.

Jenny is picking a marble from

a jar with two green marbles, six

yellow marbles, and two red

marbles.

rolling a six-sided die

arrowRight

selecting a playing card from

the hearts suit

arrowRight

spinning a spinner divided into

five equal sections

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

In a simple random sample of 1000 Americans, it was found that 61% were satisfied with the service provided by the dealer from w

BlackZzzverrR [31]

Answer:

c. smaller for the sample of Canadians, since the percentage satisfied was smaller than that for the Americans.

Correct we see that the proportion of Canadians is lower than the proportion of Americans since 0.58<0.61. And the reason why is by the percentage.

Step-by-step explanation:

Data given and notation

$X_{A}=610$ represent the number of people of America satisfied

$X_{C}=580$ represent the number of people of Canada satisfied

$n_{A}=1000$ sample of Americans selected

$n_{C}=1000$ sample of Canadians selected

$p_{A}=\frac{610}{1000}=0.61$ represent the proportion of American people satisifed

$p_{C}=\frac{580}{1000}=0.58$ represent the proportion of Canadians satisfied

$\hat p$ represent the pooled estimate of p

Solution to the problem

If we analyze one by one the optiosn we have this:

a. larger for the Canadians because the sample size is smaller.

False, both have the same the same sample size of 1000 and the proportion of Canadians is less than the proportion of Americans.

b. smaller for the sample of Canadians, since the population of Canada is less than half that of the United States. Hence, the sample is a larger proportion of the population.

False, the proportion of Canada is not less than half of the United States since the difference is just 0.61-0.58=0.03

c. smaller for the sample of Canadians, since the percentage satisfied was smaller than that for the Americans.

Correct we see that the proportion of Canadians is lower than the proportion of Americans since 0.58<0.61. And the reason why is by the percentage.

d. larger for the Canadians, since Canadian citizens are more widely dispersed throughout their country than American citizens are in the United States. Hence, Canadians have more variable views.

False the porportion of Canadians 0.58 is NOT higher than the proportion of Americans satisfied 0.61 since 0.58 is not less than 0.61.

5 0

3 years ago

Read 2 more answers

Help <br> other do divide

vovikov84 [41]

Can you say exactly what your asking for.

3 0

3 years ago

Solve for v please? use additive property

Ahat [919]

V= -13
-7=v+6
-6 -6
-13=v

7 0

3 years ago