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pentagon [3]
3 years ago
6

A bank loans ​$2100 to a business at 7​% for 3 months. How much simple interest will the bank​ earn?

Mathematics
1 answer:
Natali [406]3 years ago
7 0

Answer:

  $36.75

Step-by-step explanation:

The amount of interest payable is given by ...

  I = Prt = $2100·0.07·(3/12) . . . . . . . . t is in years

  I = $36.75

The bank earned $36.75 on the loan.

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Step-by-step explanation:

y^5*y^3=y^{5+3}=y^8

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Read 2 more answers
An individual who has automobile insurance from a certain company is randomly selected. Let X be the number of moving violations
grin007 [14]

Answer:

(X)      0        1          2          3         4

P(X)   0.17   0.23    0.27    0.24    0.09

F(x)    0.17    0.04    0.65    0.91      1      

Step-by-step explanation:  

Given that;

(X)      0        1          2          3         4

P(X)   0.17   0.23    0.27    0.24    0.09

cumulative distribution function can be calculated by;  be cumulatively up the value of p(x) with the values before it;

so

x      F(x)

0     P(X = 0) = 0.17

1       P(X = 0) + P(X = 1) = 0.17 + 0.23 = 0.4

2      P(X = 0) + P(X = 1) + P(X = 2) = 0.17 + 0.23 + 0.27 = 0.65

3      P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.17 + 0.23 + 0.27 + 0.24 = 0.91

4      P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.17 + 0.23 + 0.27 + 0.24 + 0.09 = 1

Therefore, cumulative distribution function f(x) is;

(X)      0        1          2          3         4

P(X)   0.17   0.23    0.27    0.24    0.09

F(x)    0.17    0.04    0.65    0.91      1      

8 0
2 years ago
Listed below are prices in dollars for one night at different hotels in a certain region. Find the​ range, variance, and standar
Artyom0805 [142]

Answer:

Range = 115$

Standard Deviation = 43.76$

Variance = 1915.142$

Option  A) The measures of variation are not very useful because when searching for a​ room, low​ prices, location, and good accommodations are more important than the amount of variation in the area.

Step-by-step explanation:

We are given the data for  prices in dollars for one night at different hotels in a certain region.

234, 160, 119, 131, 218, 207, 146, 141        

Range:

Sorted data: 119, 131, 141, 146, 160, 207, 218, 234

\text{Range} = 234-119 = 115\$

Standard Deviation:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1356}{8} = 169.5

Sum of squares of differences = 4160.25 + 90.25 + 2550.25 + 1482.25 +  2352.25 + 1406.25 + 552.25 + 812.25 = 13406

\sigma = \sqrt{\dfrac{13406}{7}} = 43.76\$

Variance =

\sigma^2 = 1915.142\$

Measure of variance for someone searching for room:

Option  A) The measures of variation are not very useful because when searching for a​ room, low​ prices, location, and good accommodations are more important than the amount of variation in the area.

5 0
3 years ago
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