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3 years ago

11

A manager samples the receipts of every fifth person who goes through the line. Out of 50 people, 4 had a mispriced item. If 1,6

00 people go to this store each day, how many people would you expect to have a mispriced item?

1 answer:

rusak2 [61]3 years ago

4 0

Answer:

128

Step-by-step explanation:

There are 32 50's in 1600.

So 32x4=128

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Ling worked three more hours on Tuesday than she did on Monday. On Wednesday, she worked one hour more than twice the number of

Nina [5.8K]

Let x be the number of hours ling work on monday.

We know that she worked three more hours on tuesday that in monday, this can be express as :

$x+3$

We also know that in wednesday she worked on more hour than twice the number on mondays, this can be expressed as:

$2x+1$

The total number of hours she worked this three days in two more than five the number of hours she worked on monday, this can be express as :

$5x+2\\$

Now , once we have all the expressions we add the expressions of the days and equate them to the total

$x+(x+3)+(2x+1)=5x+2$

Now we solve the equation

$x+(x+3)+(2x+1)=5x+2\\x+x+3+2x+1=5x+2\\4x+4=5x+2\\5x-4x=4-2\\x=2$

Therefore , she worked 2 hours on monday.

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joja [24]

$171.45

135 x 0.27 + 135

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Find the slope of the line.<br><br> A.) 3/2<br> B.) - 2/3<br> C.) - 3/4<br> D.) 4/3

Julli [10]

Option C is your answer.

It's a negative sloping line which eliminates A and D. Then you just do the whole rise-over-run deal to get the 3/4 part.

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The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct

lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

$y=\dfrac{x}{1+x}$

Point $P\left(1,\dfrac{1}{2}\right)$ lies on the curve.

Point $Q\left(x,\dfrac{x}{1+x}\right)$ is an arbitrary point on the curve.

The slope of the secant line PQ is

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]$

1. If x=0.5, then the slope is

$\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333$

2. If x=0.9, then the slope is

$\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632$

3. If x=0.99, then the slope is

$\dfrac{1}{2(0.99+1)}=\dfrac{1}{3.98}\approx 0.2513$

4. If x=0.999, then the slope is

$\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501$

5. If x=1.5, then the slope is

$\dfrac{1}{2(1.5+1)}=\dfrac{1}{5}\approx 0.2$

6. If x=1.1, then the slope is

$\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381$

7. If x=1.01, then the slope is

$\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488$

8. If x=1.001, then the slope is

$\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499$

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4 A figure is graphed on a coordinate grid as shown.

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Answer:

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Step-by-step explanation:

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