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nadya68 [22]
3 years ago
7

The price of a computer component is decreasing at a rate of 13​% per year. State whether this decrease is linear or exponential

. If the component costs ​$90 ​today, what will it cost in three​ year
Mathematics
1 answer:
nika2105 [10]3 years ago
3 0

Answer:

$59.27

Step-by-step explanation:

It is exponential decrease.

After each year goes by its worth 100 - 0.13 = 0.87 of the previous value.

The equation of decrease  is V = 90(0.87)^t   where t = the number of years.

So after 3 years it is worth  90(0.87)^3

= $59.27.

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Several years​ ago, 38​% of parents who had children in grades​ K-12 were satisfied with the quality of education the students r
Colt1911 [192]

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0.428 - 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.3995

0.428 + 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.4564

We are confident that the true proportion of people satisfied with the quality of education the students receive is between (0.3995, 0.4564), since the lower value for this confidence level is higher than 0.38 we have enough evidence to conclude that the parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

For this case we are interesting in the parameter of the true proportion of people satisfied with the quality of education the students receive

The confidence level is given 95%, the significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical values are:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The estimated proportion of people satisfied with the quality of education the students receive is given by:

\hat p =\frac{499}{1165}= 0.428

The confidence interval for the proportion if interest is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

And replacing the info given we got:

0.428 - 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.3995

0.428 + 1.96\sqrt{\frac{0.428(1-0.428)}{1165}}=0.4564

We are confident that the true proportion of people satisfied with the quality of education the students receive is between (0.3995, 0.4564), since the lower value for this confidence level is higher than 0.38 we have enough evidence to conclude that the parents' attitudes toward the quality of education have changed.

8 0
3 years ago
In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
Tcecarenko [31]

Answer:

a. 0.7291

b. 0.9968

c. 0.7259

Step-by-step explanation:

a. np and n(1-p) can be calculated as:

np=23\times 0.48\\\\=11.04\\\\n(1-p)=23(1-0.52)\\\\=11.96

#Both np and np(1-p) are greater than 5, hence, normal approximation is most appropriate:

\mu_x=11.04\\\\\sigma^2=np(1-p)=0.48\times 0.52\times 23=5.7408

#Define Y:

Y~(11.04,5.7408)

P(X\leq 12)\approx P(Y\leq 12.5)\\\\P(Z\leq \frac{(12.5-11.04)}{\sqrt{5.7408}})=\\\\=1-0.2709\\\\=0.7291

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b. The  probability that 5 or more fish were caught.

#Using normal approximation:

P(X \geq 5) \approx P(Y \geq 4.5) = P(Z \geq\frac{ (4.5-11.04)}{\sqrt{(5.7408)}})\\\\=1-0.0032\\\\=0.9968

Hence, the probability of catching 5+ is 0.9968

c. The probability of between 5 and 12 is calculated as;

-From b above P(X\geq 5)=0.9968 and a ,P(X\leq 12)=0.7291

P(5\leq X\leq 120\approx P(4.5\leq Y\leq  12.5)\\\\=0.7291-(1-0.9968)\\\\=0.7259

Hence, the probability of between 5 and 12 is 0.7259

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3 years ago
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