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3 years ago

Landon invested $110 in an account paying an interest rate of 4.1% compounded

Mathematics

1 answer:

krok68 [10]3 years ago

7 0

Answer:

the amount after 5 years using compound continuously is $135.03

Step-by-step explanation:

The computation of the amount after 5 years using compound continuously is as follows

= Principal × e^(rate × time period)

= $110 × e^(4.2% × 5)

= $110 × 1.227525065

= $135.03

Hence, the amount after 5 years using compound continuously is $135.03

We simply applied the above formula so that the correct value could come

And, the same is to be considered

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Add the followinf expresion b. x- 3y + 4z , y – 2x- 8z ,5x – 2y – 3z

Mariana [72]

Answer:

4x - 4y - 7z

Step-by-step explanation:

x - 3y + 4z + y - 2x - 8z + 5x - 2y - 3z =

(x - 2x + 5x) + (y - 3y - 2y) + (4z - 8z - 3z) =

4x + (- 4y) + (-7z) =

4x - 4y - 7z <===

3 0

3 years ago

Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form

Alex787 [66]

Answer:

$y=-\frac{2}{5}x-\frac{1}{5}$

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

m is the slope

b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

$m = -\frac{2}{5}$

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

$y=-\frac{2}{5}x + b$

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,1). When x of the line is -3, y of the line must be 1.
(2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: $y=-\frac{2}{5}x + b$ . $b$ is what we want, the - $-\frac{2}{5}$ is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

(-3,1). y = mx + b or $1=-\frac{2}{5} * -3 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(-3)$ . $b = -\frac{1}{5}$ .
(2,-1). y = mx + b or $-1=-\frac{2}{5} * 2 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points (-3,1) and (2,-1) is $y=-\frac{2}{5}x-\frac{1}{5}$

8 0

3 years ago

Please help me somebody! <3

Misha Larkins [42]

I believe it would be -17m-4/8

8 0

3 years ago

Read 2 more answers

Consider the two data sets below:

alina1380 [7]

Answer:

Option c) The data sets will have the same values of their interquartile range.

Explanation:

1. The values are in order: they are in increasing oder, from lowest to highest value.

2. Calculate the interquartile range.

Interquartile range, IQR, is the third quartile, Q3, less the first quartile Q1:

IQR = Q3 - Q1

To find the first and the third quartile, first find the median:

Data Set 1: 19, 25, 35, 38, 41, 49, 50, 52, 59

[19, 25, 35, 38], 41, [49, 50, 52, 59]

↑

median = 41

Data Set 2: 19, 25, 35, 38, 41, 49, 50, 52, 99

[19, 25, 35, 38] , 41, [49, 50, 52, 99]

↑

median = 41

Now find the median of each subset: the values below the median and the values above the median.

Data set 1: First quartile

[19, 25, 35, 38],

↑

Q1 = [25 + 35] / 2 = 30

Third quartile

[49, 50, 52, 59]

↑

Q3 = [50 + 52] / 2 = 51

IQR = Q3 - Q1 = 51 - 30 = 21

Data set 2: First quartile

[19, 25, 35, 38]

↑

Q1 = [25 + 35] / 2= 30

Third quartile

[49, 50, 52, 99]

↑

Q3 = [52 + 50]/2 = 51

IQR = 51 - 30 = 21

Thus, it is shown that the data sets have will have the same values for the interquartile range: IQR = 21. (option c)

This happens because replacing one extreme value (in this case the maximum value) by other extreme value does not affect the median.

An outlier will change the range because the range is the maximum value less the minimum value.

5 0

3 years ago

Using the binomial theorem, expand the expression (4 + y) to the power of 4

Flauer [41]

Answer:

When raised to the power of 4, the binomial (4 + y) expands to:

$y^4 + 16y^3 + 96y^2 + 256y + 256$

Step-by-step explanation:

$(4 + y)^4 \\= (y^2 + 8y + 16)^2\\= y^4 + 8y^3 + 16y^2 + 8y^3 + 64y^2 + 128y + 16y^2 + 128y + 256\\= y^4 + 16y^3 + 96y^2 + 256y + 256$

8 0

3 years ago