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motikmotik
3 years ago
15

By definition, simple interest is:

Mathematics
1 answer:
yan [13]3 years ago
3 0

Answer:

interest paid on principal only.

Step-by-step explanation:

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Find the distance between V(4,4) and X(5,8).<br> Round to the nearest tenth, if necessary.
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Answer:

(x_{2} - x_{1})^{2} +  (y_{2} - y_{1})^{2} = c^{2}

V(x_{1}, y_{1}) = V(4, 4)\\X(x_{2}, y_{2}) = V(5, 8)\\

(5-4)^{2} + (8-4)^{2} = c^{2}\\1^{2} + 4^{2} = c^{2}\\

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Step-by-step explanation:

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Read 2 more answers
if the mean of a population is 250 and its standard de- viation is 20, approximately what proportion of obser- vations is in the
Anton [14]

Observed proportions (relative frequencies) from a sample are statistics that can be used to estimate probabilities.

a) Thre is 99.73% proportion of observations is in between 190 and 310 .

b)there is 95.44% proportion of observations is in between 210 and 290 .

We have given that,

Mean of population, μ = 250

standard deviations, σ = 20

we have to calculate proportion of observations in the given interval pair.

a) pair of values are 190 and 310

Using Z-score formula,

Z =( x - μ) /σ

=> Z = (190 - 250) / 20 = -3

Similarly, Z = (310 - 250)/20 = 3

So, -3 < z < 3 and from Z score we found p-value

P(190<X<310) = P( 190-u/sd < X-u/sd < 310-u/sd)

= P( X< 310 ) - P(X<190) = P(Z < 3) - P( Z< -3 )

=0.99865 - 0.00135 0.9973 = 99.73%

b) pair is 210 and 290

Z-Score = (210-250)/20 = -2

Z-score = (290-250)/20 = 2

so, -2 < z < 2

From Z score we found p-value

P(210<X<290) = P(X<290) - P(X<210)

= P(Z< 2) - P(Z< -2)

= 0.9772 - 0.0228 = 0.9544 = 95.44%

So, the required proportion are 99.73% and 95.44% proportion of observations in each given pair.

To learn more about Proportion value , refer:

brainly.com/question/19131394

#SPJ4

Complete question:

If the mean of a population is 250 and its standard deviation is 20, approximately what proportion of observations is in the interval between each pair of values? a 190 and 310 b. 210 and 290.

6 0
1 year ago
Kathy deposits $25 into an investment account with an
Oksi-84 [34.3K]

Answer:

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3 years ago
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