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anyanavicka [17]
3 years ago
14

The bird population in Louisiana is decreasing at a rate of 15% per day due to the oil spill. If there are 12,000 birds today, h

ow many will there be three weeks from now?
Mathematics
1 answer:
tiny-mole [99]3 years ago
8 0
<h3>Answer: 395 birds</h3>

==============================================

Work Shown:

The formula to use is

A = P*(1+r)^x

where,

P = 12000 is the initial population

r = -0.15 represents the growth rate in decimal form; which is really a decay/decrease due to r being negative

x = 21 days have passed by (3 weeks = 3*7 = 21 days)

These values lead to...

A = P*(1+r)^x

A = 12000*(1+(-0.15))^21

A = 12000*(1-0.15)^21

A = 12000*(0.85)^21

A = 12000*0.03294560142183

A = 395.34721706196

A = 395

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2 3/4 in.

Step-by-step explanation:

66/6 = 11

11(1/4) = 11/4

11/4 = 2 3/4

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A survey of 1,562 randomly selected adults showed that 522 of them have heard of a new electronic reader. The accompanying techn
tester [92]

Answer:

a) We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:  

Null hypothesis:p=0.35  

Alternative hypothesis:p \neq 0.35  

And is a two tailed test

b) z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326  

c) p_v =2*P(z  

d) Null hypothesis:p=0.35  

e) Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

Step-by-step explanation:

Information provided

n=1562 represent the random sample selected

X=522 represent the people who have heard of a new electronic reader

\hat p=\frac{522}{1562}=0.334 estimated proportion of people who have heard of a new electronic reader

p_o=0.35 is the value to verify

\alpha=0.05 represent the significance level

z would represent the statistic

p_v represent the p value

Part a

We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:  

Null hypothesis:p=0.35  

Alternative hypothesis:p \neq 0.35  

And is a two tailed test

Part b

The statistic for this case is given :

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Replacing the info given we got:

z=\frac{0.334 -0.35}{\sqrt{\frac{0.35(1-0.35)}{1562}}}=-1.326  

Part c

We can calculate the p value using the laternative hypothesis with the following probability:

p_v =2*P(z  

Part d

The null hypothesis for this case would be:

Null hypothesis:p=0.35  

Part e

The best conclusion for this case would be:

Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

5 0
3 years ago
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