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Zarrin [17]
3 years ago
14

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

Mathematics
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
I am Lyosha [343]

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

5 0
3 years ago
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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
melamori03 [73]

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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3 years ago
Classify the following triangle as acute,obtuse, or right
IRISSAK [1]
It is an obtuse triangle
4 0
3 years ago
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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

8 0
3 years ago
How do I find 2/3 of 12 slices of pizza eaten ​
MariettaO [177]

Multiply 12 slices by 2/3:

12 x 2/3 = (12 x 2) /3 = 24/3 = 8

The answer is 8

7 0
3 years ago
Read 2 more answers
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