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3 years ago

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A saleswoman is paid $80 a day plus $6 for each sale she makes. She wants to earn more than $150 today. Write an inequality that

represents the number of sales, x, the saleswoman must make today to earn more than $150. What is the least number of sales she must make to reach her sales goal

1 answer:

Zarrin [17]3 years ago

6 0

Answer:

12

Step-by-step explanation:

Given that:

Daily wages paid to a saleswoman = $80

Wages paid for each sale made by her = $6

Target of today's earnings = more than $150

To find:

Least number of sales to be made so that the total earnings today is more than $150.

Solution:

Number of sales for today = $x$

Wages for one sale = $6

Wages for $x$ number of sales = $6 $x$

Total earnings = Per day sales + Wages for $x$ number of sales > $150

$80 + 6x > 150$

we need to solve for above inequality to find the least number of sales to be made to meet the goal.

Subtracting 80 from both sides:

$\Rightarrow 6x>70\\$

Dividing both sides by 6:

$\Rightarrow x > 11$

Therefore, <em>at least 12 sales</em> must be made.

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Answer:

X=-4, y=-7

Step-by-step explanation:

6x-5y=11. 1st equation

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6x-35x-105=11. Solve by distributative property

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Answer:

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Step-by-step explanation:

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During a week-long dry spell, the water level in a pond decreased by 4 in. Per day, except for two days when it decreased by hal

Helga [31]

Answer:

24 inches.

Step-by-step explanation:

Given:

During a week the water level in a pond decreased by 4 in. Per day,

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Question asked:

How much did the water level decrease in the pond in one week?

Solution:

As we know, there are 7 days in a week.

<u>For 5 days,</u>

Each day water level decreased by = 4 inches

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<u>For exceptional 2 days:</u>

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3 years ago

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nadezda [96]

Answer:

2

Step-by-step explanation:

Order of operations is not as applicable in this situation because we're just dealing with addition and subtraction. This is a simple case of plugging in values for the given expression.

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2 years ago

A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. Th

Olegator [25]

Answer:

(a) 74553

(b) 172120

(c) 234802

Step-by-step explanation:

Given

$y = \frac{340110}{1 + 377e^{-0.259t}}$

Solving (a): 1998

Year 1998 means that:

$t =1998 - 1980$

$t =18$

So, we have:

$y = \frac{340110}{1 + 377e^{-0.259*18}}$

$y = \frac{340110}{1 + 377e^{-4.662}}$

$y = \frac{340110}{1 + 3.562}$

$y = \frac{340110}{4.562}$

$y = 74553$ --- approximated

Solving (b): 2003

Year 2003 means that:

$t = 2003 - 1980$

$t =23$

So, we have:

$y = \frac{340110}{1 + 377e^{-0.259*23}}$

$y = \frac{340110}{1 + 377e^{-5.957}}$

$y = \frac{340110}{1 + 0.976}$

$y = \frac{340110}{1.976}$

$y = 172120$ --- approximated

Solving (c): 2006

Year 2006 means that:

$t = 2006 - 1980$

$t =26$

So, we have:

$y = \frac{340110}{1 + 377e^{-0.259*26}}$

$y = \frac{340110}{1 + 377e^{-6.734}}$

$y = \frac{340110}{1 + 0.4485}$

$y = \frac{340110}{1.4485}$

$y = 234802$ --- approximated

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