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3 years ago

50 - z = 10 What is the coefficient of the unknown in the expression above

Mathematics

2 answers:

shepuryov [24]3 years ago

4 0

50 - z = 10
-50 -50
_________
-z = -40

The coefficient of the unknown
expression above is -1

stira [4]3 years ago

3 0

The coefficient is 1 because a coefficient is the number attached to the variable. In this case the variable(z) does not have a visible number but there is always a number if the number is not visible it is a 1. So the coefficient is one.

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Justify solution -1/2(5x-8)=7 5x-8=-14 5x=-6 x=-6/5

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Answer:

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-1/2(5x-8)=7

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Read 2 more answers

A service station checks Mr. Smith's radiator and finds it contains only 40% antifreeze. If the radiator holds 20 quarts and is

storchak [24]

Answer:

8 quarts

Step-by-step explanation:

Step One: Figure out how much 40% is by making an equation:

Since we are looking for a percentage, we use cross multiplication to find X.

$\frac{x}{20} = \frac{40}{100}$

Step Two: multiply 20 by 40 to start cross multiplying:

20 times 40 is 800, so replace the 40 by 800 in the equation.

$\frac{x}{20} = \frac{800}{100}$

Step Three: divide 800 by 100, and we will have X:

800/100=8

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I hope this helps!

pls give thx and brainliest :D

7 0

3 years ago

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Blababa [14]

Answer:

2,000

4,000

7,000

put points on (2, 2,000) - (4, 4,000) - and (7, 7,000)

Step-by-step explanation:

6 0

2 years ago

Assume we need to estimate the mean of a normally-distributed population with great accuracy. Specifically, for significance lev

bulgar [2K]

Answer:

$n = 2662.56\sigma^2$

Step-by-step explanation:

We are given the following in the question:

Significance level = 0.01

Width of interval = 0.1

Population variance = $\sigma^2$

We have to find the sample size so that the width of the confidence interval is no larger than 0.1

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58$

Formula for sample size:

$n = \displaystyle\frac{z^2\sigma^2}{E^2}$

where E is the margin of error. Since the confidence interval width is 0.1,

$E = 0.05$

Putting these values in the equation:

$n = \displaystyle\frac{(2.58)^2\sigma^2}{(0.05)^2} = 2662.56\sigma^2$

So, the above expression helps us to calculate the sample size so that the width of the confidence interval is no larger than 0.1 for different sample variances.

4 0

4 years ago