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BartSMP [9]
3 years ago
15

Solve for x.

Mathematics
2 answers:
JulijaS [17]3 years ago
4 0

Answer:

x = 2281/2630 = 0.867

AlladinOne [14]3 years ago
4 0

Answer:

-2

Step-by-step explanation:

Just took the test

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Frank is lending $1,000 to Sarah for two years. Frank and Sarah agree that Frank should earn a real return of 4 percent per year
Montano1993 [528]

Answer:

nominal rate of interest is 31.7 %

Step-by-step explanation:

given data

payment = $1000

time = 2 year

rate = 4%

CPI = 100

CPI final yer = 127.7

to find out

nominal rate of interest

solution

we know nominal rate of interest formula that is

nominal rate of interest = real interest rate + inflation rate    .........1

so here inflation rate is express as

inflation rate = ( CPI final year - CPI )  / CPI  × 100    ...........2

put here value

inflation rate = ( 127.7 - 100 )  / 100  × 100

inflation rate is 27.7 %

so from equation 1

nominal rate of interest = real interest rate + inflation rate

nominal rate of interest = 4% + 27.7%

nominal rate of interest is 31.7 %

7 0
3 years ago
Can someone help me out here ??
Vaselesa [24]

Answer:

6.04 inches

Step-by-step explanation:

1 in. = 100 Ft.

604 Ft / 100Ft = 6.04

5 0
2 years ago
Read 2 more answers
Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and
Irina18 [472]

Answer:

0.918 is the probability that the sample average sediment density is at most 3.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.80

Standard Deviation, σ = 0.85

Sample size,n = 35

We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

Standard error due to sampling:

=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437

P(sample average sediment density is at most 3.00)

P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)

Calculation the value from standard normal z table, we have,  

P(x \leq 3.00) = 0.918

0.918 is the probability that the sample average sediment density is at most 3.00

4 0
3 years ago
How many hours are in 186 minutes? Express this as a mixed number and decimal
Solnce55 [7]
3.1 as a decimal. And it as a mixed number wait a sec.......3 1/10
Hope I could help! :)
6 0
3 years ago
n automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally di
Paraphin [41]

Answer:

0.3557 = 35.57% probability that one selected subcomponent is longer than 118 cm.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of 116 cm and a standard deviation of 5.4 cm.

This means that \mu = 116, \sigma = 5.4

Find the probability that one selected subcomponent is longer than 118 cm.

This is 1 subtracted by the pvalue of Z when X = 118. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{118 - 116}{5.4}

Z = 0.37

Z = 0.37 has a pvalue of 0.6443

1 - 0.6443 = 0.3557

0.3557 = 35.57% probability that one selected subcomponent is longer than 118 cm.

8 0
3 years ago
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