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3 years ago

Fidel has a rare coin worth \$550$550dollar sign, 550. Each decade, the coin's value increases by 10\%10%10, percent.

Mathematics

2 answers:

FinnZ [79.3K]3 years ago

8 0

Answer:

Option B - $550(1+0.1)^6$

Step-by-step explanation:

Given : Fidel has a rare coin worth $550. Each decade, the coin's value increases by 10%.

To find : Which expression gives the coin's value, 6 decades from now?

Solution :

The given situation represents the exponential function,

$y=a(1+r)^n$

where, a is the initial value i.e. a=$550

r is the increased rate r=10%

n is the time i.e. n=6

Substitute the value in the formula,

$y=550(1+10\%)^6$

$y=550(1+0.1)^6$

Therefore, option B is correct.

lesya [120]3 years ago

6 0

Answer: c

Step-by-step explanation:

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Komok [63]

Answer:

99% confidence interval is wider as compared to the 80% confidence interval.

Step-by-step explanation:

We are given that a magazine provided results from a poll of 500 adults who were asked to identify their favorite pie.

Among the 500 respondents, 14% chose chocolate pie, and the margin of error was given as plus or minus ±3 percentage points.

The pivotal quantity for the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of adults who chose chocolate pie = 14%

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Now, the 99% confidence interval for p = $\hat p \pm Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

Here, $\alpha$ = 1% so $(\frac{\alpha}{2})$ = 0.5%. So, the critical value of z at 0.5% significance level is 2.5758.

Also, Margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$ = 0.03 for 99% interval.

So, 99% confidence interval for p = $0.14 \pm2.5758 \times \sqrt{\frac{0.14(1-0.14)}{500} }$

= [0.14 - 0.03 , 0.14 + 0.03]

= [0.11 , 0.17]

Similarly, 80% confidence interval for p = $0.14 \pm 1.2816 \times \sqrt{\frac{0.14(1-0.14)}{500} }$

Here, $\alpha$ = 20% so $(\frac{\alpha}{2})$ = 10%. So, the critical value of z at 10% significance level is 1.2816.

Also, Margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$ = 0.02 for 80% interval.

So, 80% confidence interval for p = [0.14 - 0.02 , 0.14 + 0.02]

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Now, as we can clearly see that 99% confidence interval is wider as compared to 80% confidence interval. This is because more the confidence level wider is the confidence interval and we are more confident about true population parameter.

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