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3 years ago

PLEASE HELP 15 POINTS,

Mathematics

1 answer:

kenny6666 [7]3 years ago

7 0

Step-by-step explanation:

I take gas consumption as example.

Consider the followings,

Let M be the gas meter reading per month

A be the fuel adjustment fee

B be monthly maintenance fee

Y be unit cost per Joule

Let X be monthly bill

( 1 meter reading unit = 48 unit of Joule )

Monthly bill payment :

X = Mx48xY + A + B

in function notation, the monthly bill is f(M, Y, A, C)

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Put together, Dulcina and Tremaine have 129 total matchbooks. Tremaine's collection has 39 fewer matchbooks in it than Dulcina's

Anit [1.1K]

Answer:

Let the Dulcina's collection be 'x'

Let the Tremaine collection be 'x-39'

x + x - 39 =129

2x = 129 +39

2x = 168

x = 168/2

x = 84

Dulcina's collection = x = 84

Tremaine's collection = x - 39 = 84 - 39 = 45

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3 years ago

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Step-by-step explanation:

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3 years ago

Read 2 more answers

(K3+2k2-82k-28)/(k+10)

Ne4ueva [31]

So for this, we will be using synthetic division. To set it up, have the equation so that the divisor is -10 (since that is the solution of k + 10 = 0) and the dividend are the coefficients. Our equation will look as such:

(Note that synthetic division can only be used when the divisor is a 1st degree binomial)

-10 | 1 + 2 - 82 - 28
---------------------------

Now firstly, drop the 1:

-10 | 1 + 2 - 82 - 28
↓
-------------------------
1

Next, you are going to multiply -10 and 1, and then combine the product with 2.

-10 | 1 + 2 - 82 - 28
↓ - 10
-------------------------
1 - 8

Next, multiply -10 and -8, then combine the product with -82:

-10 | 1 + 2 - 82 - 28
↓ -10 + 80
-------------------------
1 - 8 - 2

Next, multiply -10 and -2, then combine the product with -28:

-10 | 1 + 2 - 82 - 28
↓ -10 + 80 + 20
-------------------------
1 - 8 - 2 - 8

Now, since we know that the degree of the dividend is 3, this means that the degree of the quotient is 2. Using this, the first 3 terms are k^2, k, and the constant, or in this case k² - 8k - 2. Now what about the last coefficient -8? Well this is our remainder, and will be written as -8/(k + 10).

Putting it together, the quotient is $k^2-8k-2-\frac{8}{k+10}$ 

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3 years ago

My Notes A large manufacturing plant uses lightbulbs with lifetimes that are normally distributed with a mean of 1600 hours and

Evgen [1.6K]

Answer:

The bulbs should be replaced each 1436.9 hours.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 1600, \sigma = 70$