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3 years ago

A MACHINE WORKED 5 HOURS USING 4 KILOWATTS OF ELECTRICITY. HOW MUCH WOULD MACHINE HAVE USED IN KILOWATT EACH HOUR WORKED

1 answer:

5 0

4 kilowatt is a power, not work done.

If the machine works using power at the rate of 4 kilowatt, then it is using 4 kilowatt each hour.

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Hey anyone else bored

german

Answer:

same

Step-by-step explanation:

im actually in class rn ot sucks

3 0

2 years ago

Read 2 more answers

Please help!!! 10 points up!

Dimas [21]

Statements are all TRUE

sinA = $\frac{a}{c}$ and cosB = $\frac{a}{c}$

cscA = 1/sinA = 1/(a/c) = $\frac{c}{a}$ and secB = 1/cosB = 1/(a/c) = $\frac{c}{a}$

tanA = $\frac{a}{b}$ and cotB= 1/tanB = 1/ (b/a) = $\frac{a}{b}$

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3 years ago

2x+1=3(y-1)<br> 6(2y-1)=5x<br> Find values of x and y

olchik [2.2K]

Answer:

x= -10/3

y= -8/9

Step-by-step explanation:

2x+1=3(y−1);6(2y−1)=5x

Rewrite equations:

2x+1=3y−3;12y−6=5x

Step: Solve2x+1=3y−3for x:

2x+1=3y−3

2x+1+−1=3y−3+−1(Add -1 to both sides)

2x=3y−4

2x2=3y−42(Divide both sides by 2)

x=32y−2

Step: Substitute32y−2forxin12y−6=5x:

12y−6=5x

12y−6=5(32y−2)

12y−6=152y−10(Simplify both sides of the equation)

12y−6+−152y=152y−10+−152y(Add (-15)/2y to both sides)

92y−6=−10

92y−6+6=−10+6(Add 6 to both sides)

92y=−4

92y92=−492(Divide both sides by 9/2)

y=−89

Step: Substitute−89foryinx=32y−2:

x=32y−2

x=32(−89)−2

x=−103(Simplify both sides of the equation)

4 0

3 years ago

Read 2 more answers

The demand for a daily newspaper at a newsstand at a busy intersection is known to be normally distributed with a mean of 150 an

AURORKA [14]

Answer:

171 newspapers.

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:

$\mu = 150, \sigma = 25$