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3 years ago

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Find the simple interest on a $3,219.00 principal, deposited for six years at a rate of 1.51%.

1 answer:

Karo-lina-s [1.5K]3 years ago

8 0

Answer:

s.i = P. T. R /100

= 3219 . 6 . 1.51 /100

= 29,164.14 / 100

= 291.6414

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What is the value of x for which (8-x)^2=x^2

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Answer:

x = 4

Step-by-step explanation:

${x}^{2} = {(8 - x)}^{2}$

$= > {x}^{2} = {x}^{2} - 16x + 64$

$= > {x}^{2} - {x}^{2} + 16x = 64$

$= > 16x = 64$

$= > x = \frac{64}{16} = 4$

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Pyramid a has a lateral edge of 10 and a height of 8. Pyramid B has a lateral edges of length 15 and height 12. Both A and B are

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Answer:

Pyramid A is 10*10*8/3 which is 800/3 and Pyramid B is 15*15*12/3 which is 2700/3, so the ratio would simply be 8:27 uwu

Step-by-step explanation:

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Classify the following triangle as acute,obtuse, or right

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It is an obtuse triangle

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The estimated daily living costs for an executive traveling to various major cities follow. The estimates include a single room

Alexandra [31]

Answer:

$\bar x = 260.1615$

$\sigma = 70.69$

The confidence interval of standard deviation is: $53.76$ to $103.25$

Step-by-step explanation:

Given

$n =20$

See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

$\bar x = 260.1615$

$\bar x = 260.16$

Solving (b): The standard deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

$\sigma = \sqrt{4996.78}$

$\sigma = 70.69$ --- approximated

Solving (c): 95% confidence interval of standard deviation

We have:

$c =0.95$

So:

$\alpha = 1 -c$

$\alpha = 1 -0.95$

$\alpha = 0.05$

Calculate the degree of freedom (df)

$df = n -1$

$df = 20 -1$

$df = 19$

Determine the critical value at row $df = 19$ and columns $\frac{\alpha}{2}$ and $1 -\frac{\alpha}{2}$

So, we have:

$X^2_{0.025} = 32.852$ ---- at $\frac{\alpha}{2}$

$X^2_{0.975} = 8.907$ --- at $1 -\frac{\alpha}{2}$

So, the confidence interval of the standard deviation is:

$\sigma * \sqrt{\frac{n - 1}{X^2_{\alpha/2} }$ to $\sigma * \sqrt{\frac{n - 1}{X^2_{1 -\alpha/2} }$

$70.69 * \sqrt{\frac{20 - 1}{32.852}$ to $70.69 * \sqrt{\frac{20 - 1}{8.907}$

$70.69 * \sqrt{\frac{19}{32.852}$ to $70.69 * \sqrt{\frac{19}{8.907}$

$53.76$ to $103.25$

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3 years ago

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Multiply 12 slices by 2/3:

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