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3 years ago

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An electrician is hired to install outdoor lighting. The electrician claims that the relationship between the number of hours wo

rked and the total work fee is porportional. The fee for 5 hours of work is $255

1 answer:

Bingel [31]3 years ago

6 0

<h2>Explanation:</h2><h2></h2>

If we have two variables and can write a relationship between them as:

$y=kx \\ \\ \\ Where: \\ \\ \text{k is the constant of proportionality}$

Then we say that $y$ varies directly as $x$ or $y$ is directly proportional to $x$

In this case, let's say:

$x:\text{The number of hours worked} \\ \\ y:\text{Total work fee}$

Then since we know that the fee for 5 hours of work is $255, then the constant of proportionality is:

$k=\frac{255}{5} \\ \\ k=51$

Therefore, our model can be written as:

$\boxed{y=51x}$

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4) Solve for x: 4(2x - 10) = 12x + 10

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Answer:

x=-25/2

Step-by-step explanation:

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3 years ago

Read 2 more answers

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

sleet_krkn [62]

This question is incomplete, the complete question is;

If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple are written in increasing order but are not necessarily distinct.

In other words, how many 5-tuples of integers ( h, i , j , m ), are there with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1 ?

Answer:

the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Step-by-step explanation:

Given the data in the question;

Any quintuple ( h, i , j , m ), with n ≥ h ≥ i ≥ j ≥ k ≥ m ≥ 1

this can be represented as a string of ( n-1 ) vertical bars and 5 crosses.

So the positions of the crosses will indicate which 5 integers from 1 to n are indicated in the n-tuple'

Hence, the number of such quintuple is the same as the number of strings of ( n-1 ) vertical bars and 5 crosses such as;

$\left[\begin{array}{ccccc}5&+&n&-&1\\&&5\\\end{array}\right] = \left[\begin{array}{ccc}n&+&4\\&5&\\\end{array}\right]$

= [( n + 4 )! ] / [ 5!( n + 4 - 5 )! ]

= [( n + 4 )!] / [ 5!( n-1 )! ]

= [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

Therefore, the number of 5-tuples of integers from 1 through n that can be formed is [ n( n+1 ) ( n+2 ) ( n+3 ) ( n+4 ) ] / 120

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3 years ago

8) Find the endpoint Cif M is the midpoint of segment CD and M (2, 4) and D (5,7)

Elenna [48]

Answer:

8. c. (-1, -1)

9. a. (-6, -1)

b. True

Step-by-step Explanation:

8. Given the midpoint M(2, 4), and one endpoint D(5, 7) of segment CD, the coordinate pair of the other endpoint C, can be calculated as follows:

let $D(5, 7) = (x_2, y_2)$

$C(?, ?) = (x_1, y_1)$

$M(2, 4) = (\frac{x_1 + 5}{2}, \frac{y_1 + 7}{2})$

Rewrite the equation to find the coordinates of C

$2 = \frac{x_1 + 5}{2}$ and $4 = \frac{y_1 + 7}{2}$

Solve for each:

$2 = \frac{x_1 + 5}{2}$

$2*2 = \frac{x_1 + 5}{2}*2$

$4 = x_1 + 5$

$4 - 5 = x_1 + 5 - 5$

$-1 = x_1$

$x_1 = -1$

$4 = \frac{y_1 + 7}{2}$

$4*2 = \frac{y_1 + 7}{2}*2$

$8 = y_1 + 7$

$8 - 7 = y_1 + 7 - 7$

$1 = y_1$

$y_1 = 1$

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9. a.Given segment AB, with midpoint M(-4, -5), and endpoint A(-2, -9), find endpoint B as follows:

let $A(-2, -9) = (x_2, y_2)$

$B(?, ?) = (x_1, y_1)$

$M(-4, -5) = (\frac{x_1 + (-2)}{2}, \frac{y_1 + (-9)}{2})$

$-4 = \frac{x_1 - 2}{2}$ and $-5 = \frac{y_1 - 9}{2}$

Solve for each:

$-4 = \frac{x_1 - 2}{2}$

$-4*2 = \frac{x_1 - 2}{2}*2$

$-8 = x_1 - 2$

$-8 + 2 = x_1 - 2 + 2$

$-6 = x_1$

$x_1 = -6$

$-5 = \frac{y_1 - 9}{2}$

$-5*2 = \frac{y_1 - 9}{2}*2$

$-10 = y_1 - 9$

$-10 + 9 = y_1 - 9 + 9$

$-1 = y_1$

$y_1 = -1$

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b. The midpoint of a segment, is the middle of the segment. It divides the segment into two equal parts. The answer is TRUE.

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4 years ago

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Answer:

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Step-by-step explanation:

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Thus, we can say that:Yes, the random condition for finding confidence intervals is met because the values of np^ and n(1 − p^) are greater than 10.

Also, Yes, the random condition for finding confidence intervals is met because the sample size is greater than 30.

Confidence interval approach is valid if;

1) sample is a simple random sample

2) sample size is sufficiently large, which means that it includes at least 10 successes and 10 failures. In general a sample size of 30 is considered sufficient.

These two conditions are met by the sample described in the question.

So, Yes, the random conditions are met.

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3 years ago

Write the explicit formula for the geometric sequence 2, 8, 32, 128

Alja [10]

The first term, a, is 2. The common ratio, r, is 4. Thus,

a_(n+1) = 2(4)^(n).

Check: What's the first term? Let n=1. Then we get 2(4)^1, or 8. Is that correct? No.

Try this instead:

a_(n) = a_0*4^(n-1). Is this correct? Seeking the first term (n=1), does this formula produce 2? 2*4^0 = 2*1 = 2. YES.

The desired explicit formula is a_(n) = a_0*4^(n-1), where n begins at 1.

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3 years ago