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BartSMP [9]
1 year ago
10

Suppose the CPI increases this year from 208 to 221. What is the rate of inflation for this year? Round your answer to the neare

st tenth of a percent.
Mathematics
1 answer:
guajiro [1.7K]1 year ago
8 0

SOLUTION:

We are to find the rate of inflation in the given question;

The formula to be used is;

Inflation rate =

\frac{\text{CPI}_{present}-CPI_{previous}}{CPI_{previous}_{}}\text{ X 100}\begin{gathered} \frac{221-208}{208}\text{ X }\frac{100}{1} \\  \\ \frac{13}{208}X\frac{100}{1} \\  \\ 6.25\text{ \%} \\ 6.3\text{ \% (nearest tenth of a percentage)} \end{gathered}

CONCLUSION

The inflation rate of CPI that increased from 208 to 221 to the nearest tenth of a percentage is 6.3%.

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IF PASSPORT is written as RCUURQTV, then how will BOOKLET be coded?​
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3 years ago
When Rhonda was 12, her grandmother made her a quilt. Each pattern in the quilt contained 126 quares. The
Natasha_Volkova [10]

Answer:

B. The quilt consisted of 1,008.

Step-by-step explanation:

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3 years ago
On the package for a certain brand of spinach seeds there is a guarantee that, if the printed instructions are followed, of plan
kirill115 [55]

The question is incomplete. Here is the complete question.

On the package for a certain brand of spinach seeds there is a guarantee that, if the printed instructions are followed, 63% of planted seeds will germinate. A random sample of 9 seeds is chosen. If these seeds are planted according to the instructions, find the probability that 4 or 5 of them germinate. Do not round your intermiediate computations, and round your answer to three decimal places.

Answer: P(4<X<5) = 0.624

Step-by-step explanation: The probability of a seed germinate is a <u>Binomial</u> <u>Distribution</u>, i.e., a discrete probability distribution of the number of successes in a sequence of n independents experiments.

This distribution can be approximated to normal distribution by determining the values of mean and standard deviation population:

\mu=np

\sigma=\sqrt{np(1-p)}

where

n is the sample quantity

p is proportion of successes

For the spinach seeds:

Mean is

\mu=9(0.65)

\mu= 5.85

Standard deviation is

\sigma=\sqrt{9.0.65(1-0.65)}

\sigma= 1.431

Now, use

z=\frac{x-\mu}{\sigma}

to convert into a standard normal distribution.

The probability we want is between 2 values: P(4<X<5).

Therefore, we have to convert those two values:

For X = 4:

z=\frac{4-5.85}{1.431}

z = -1.29

For X = 5:

z=\frac{5-5.85}{1.431}

z = -0.59

Using z-table:

P(X>4) = 1 - P(z< -1.29) = 0.9015

P(X<5) = P(z< -0.59) = 0.2776

The probability will be

P(4<X<5) = P(X>4) - P(X<5)

P(4<X<5) = 0.9015 - 0.2776

P(4<X<5) = 0.624

The probability of 4 or 5 seeds germinate is 0.624.

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