Find the least common multiple 36, 27 and 10

The distance between city E and F is 678.35 miles and the distance between city F and G is 156.8 miles. What is the distance bet

Free_Kalibri [48]

E to F = 678.35 miles
F to G = 156.8 miles
G to H = x
Total distance = 2,457 miles
E to G = 678.35 + 156.8 miles
E to G = 835.15
G to H = 2,457 - 835.15
G to H = 1621.85 miles

8 0

3 years ago

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I need help with the question pls

Andreyy89

are there options for this question?

4 0

3 years ago

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Formulate the situation as a system of two linear equations in two variables. Be sure to state clearly the meaning of your x- an

slega [8]

Answer:

1) There were 33 $4,000 investors and 27 $8,000 investors.

2) The solution in x = 4, y = 9.

3) There were 24 nickels and 56 dimes.

Step-by-step explanation:

1) A lawyer has found 60 investors for a limited partnership to purchase an inner-city apartment building, with each contributing either $4,000 or $8,000. If the partnership raised $348,000, then how many investors contributed $4,000 and how many contributed $8,000?

I am going to say that:

x is the number of investors that contributed 4,000.

y is the number of investors that contributed 8,000.

Building the system:

There are 60 investors. So:

$x + y = 60$

In all, the partnership raised $348,000. So:

$4000x + 8000y = 348000$

I am going to simplify by 4000. So:

$x + 2y = 87$

Solving the system:

The elimination method is a method in which we can transform the system such that one variable can be canceled by addition. So:

$1)x + y = 60$

$2)x + 2y = 87$

I am going to multiply 1) by -1. So we have

$1)-x - y = -60$

$2)x + 2y = 87$

By addition, the x are going to cancel each other

$-x + x - y + 2y = -60 + 87$

$y = 27$

For x:

$x + y = 60$

$x = 60-y = 60-27 = 33$

There were 33 $4,000 investors and 27 $8,000 investors.

2) Solve the system by row-reducing the corresponding augmented matrix.

$2x + y = 17$

$x + y = 13$

This system has the following augmented matrix:

$\left[\begin{array}{ccc}2&1&17\\1&1&13\end{array}\right]$

To help the row reducing, i am going to swap the first with the second line:

$L1 L2$

So we have:

$\left[\begin{array}{ccc}1&1&13\\2&1&17\end{array}\right]$

Now, reducing the first column.

$L2 = L2 - 2L1$

So we have:

$\left[\begin{array}{ccc}1&1&13\\0&-1&-9\end{array}\right]$

Now we do:

$L2 = -L2$

And the matrix is:

$\left[\begin{array}{ccc}1&1&13\\0&1&9\end{array}\right]$

Now to reduce the second column, we do:

$L1 = L1 - L2$

$\left[\begin{array}{ccc}1&0&4\\0&1&9\end{array}\right]$ .

So the solution is:

x = 4, y = 9.

3) A jar contains 80 nickels and dimes worth $6.80. How many of each kind of coin are in the jar?

I am going to say that x is the number of nickels and y is the number of dimes.

Each nickel is worth 5 cents and each dime is worth 10 cents.

Building the system:

There are 80 coins in all:

$x + y = 80$

They are worth $6.80. So:

$0.05x + 0.10y = 6.80$

Solving the system:

$1)x + y = 80$

$2)0.05x + 0.10y = 6.80$

I am going to divide 1) by -10, so we can cancel y. So:

$1)-0.10x - 0.10y = -8$

$2)0.05x + 0.10y = 6.80$

Adding:

$-0.10x + 0.05x - 0.10y + 0.10y = -8 + 6.80$

$-0.05x = -1.2$ *(-100)

$5x = 120$

$x = \frac{120}{5}$

$x = 24$

Also

$x + y = 80$

$y = 80-x = 80-24 = 56$

There were 24 nickels and 56 dimes.

8 0

3 years ago

Wind farm configuration is a significant issue as the upwind turbines generate maximal energy while creating wakes for downwind

lord [1]

Answer:

a) [24.114,29.2858]

b) Since 30 is not in the 95% confidence interval, there is a 95% probability that 30 is not the true mean WEP

Step-by-step explanation:

a)

The 95% confidence interval is given by the interval

$\bf [ \bar x-z^*\frac{s}{\sqrt n}, \bar x+t^*\frac{s}{\sqrt n}]$

where

$\bf \bar x$ = 26.7 is the sample mean

s = 17.7 is the sample standard deviation

n = 180 is the sample size

Since the sample size is big enough, we can use the Normal N(0,1) to compute $\bf z^*$ and it would be 1.96(*) (a value such that the area under the Normal curve outside the interval [-z, z] is 5% (0.05))

and our 95% confidence interval is

$\bf [26.7-1.96*\frac{17.7}{\sqrt{180}}, 26.7+1.96*\frac{17.7}{\sqrt{180}}]=\boxed{[24.114,29.2858]}$

(*)

This value can be computed in Excel with

<em>NORMINV(1-0.025,0,1)</em>

and in OpenOffice Calc with

<em>NORMINV(1-0.025;0;1)</em>

b)

Since 30 is not in the 95% confidence interval, there is a 95% probability that 30 is not the true mean WEP

7 0

3 years ago

Find the slope of a line that is perpendicular to a line that passes through the points (7,1) and (-14,4)

d1i1m1o1n [39]

The slope of the line is -1/7, so the perpendicular line has a slope of 7

7 0

3 years ago