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DanielleElmas [232]
3 years ago
9

Choose the equation and the inequality needed to answer this question.

Mathematics
1 answer:
Flauer [41]3 years ago
5 0

Answer:

Actually, what you said you have so far is not correct.  The 2 correct answers are the 1st one (x + y = 15) and the 5th one (15x + 10y > 180)

Step-by-step explanation:

If tutoring French is x hours and scooping ice cream is y hours and he is going to work 15 hours for sure doing both, then we can add them together to get that x hours + y hours = 15 hours, or put simply:  x + y = 15.

Now we are going to throw in the added fun of the money he makes doing each.  The thing to realize here is that we can only add like terms.  So looking at the equation above, we have x hours of tutoring and y hours of scooping, so if we want to add them, we will add those number of hours together to get the total number of hours he worked, which we know to be 15.  The same goes for money.  If we add money earned from tutoring to money earned from scooping, we need that to be greater than the money he wants to earn which is 180 at least.  Because he wants to earn MORE than $180. we use the ">" sign.  Since he earns $15 an hour tutoring, that expression is $15x; since he earns $10 an hour scooping, that expression is $10y.  Now add them together (and you CAN because they are both expressions relating dollars to dollars) and set the sum > $180:

$15x + $10y > $180.  That's why your answer is not correct.  Use mine (with the understanding that you care about why yours is wrong and mine is correct) and you'll be fine.

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If each angle in a decagon is the same measure, what is that measure? 18° 180° 144° 72°
pishuonlain [190]

Answer:

Each Angle of Decagon = 144 degrees

Step-by-step explanation:

The interior angles of a decagon add up to 1440

If they are of the same measure, then every angle will be equal to:

Each Angle of Decagon = 1440/10

Each Angle of Decagon = 144 degrees

5 0
3 years ago
All bags entering a research facility are screened. The screening process is not perfect so that 77% of the bags that contain fo
garik1379 [7]

Answer:

(1) 0.5933

(2) 0.8955

Step-by-step explanation:

We are given that all bags entering a research facility are screened.

Let Probability that bags entering the building contains forbidden material,

 P(F) = 0.69

Probability that bags entering the building does not contains forbidden material,   P(NF) = 1 - 0.69 = 0.31

Let event A = alarm gets triggered

Probability that alarm gets trigger given the bags contain forbidden material, P(A/F) = 0.77

Probability that alarm gets trigger given the bags does not contain forbidden material, P(A/NF) = 0.20

(1) Probability that a bag triggers the alarm, P(A) ;

         P(A) = P(F) * P(A/F) + P(NF) * P(A/NF)

                 = (0.69 * 0.77) + (0.31 * 0.20) = 0.5313 + 0.062

                 = 0.5933

Therefore, probability that a bag triggers the alarm is 0.5933 .

(2) Probability that a bag that triggers the alarm will actually contain forbidden material is given by P(F/A) ;

Using Bayes' Theorem;

    P(F/A) = \frac{P(F) * P(A/F)}{P(F) * P(A/F) + P(NF) *P(A/NF)} = \frac{0.69*0.77}{0.69*0.77+0.31*0.20} = \frac{0.5313}{0.5933}

               = 0.8955

6 0
3 years ago
if joe and jim combined their hourly pay,they would have $16. jims hourly pay is $1.50 more than joes. find each bous hourly pay
evablogger [386]
Joe : x
jim : y

x+y=16
y=1.50+x

x + 1.50 + x = 16

2x = 14.5

x = 7.25

7.25 + y = 16

y= 8.75

Joe : $7.25
Jim : $8.75
6 0
3 years ago
This problem models pollution effects in the Great Lakes. We assume pollutants are flowing into a lake at a constant rate of I k
Alenkasestr [34]

Answer:

C(t) = I/F [1 - e^(-Ft/V) ] + C₀e^(-Ft/V)

as t = 0 ; C(t) = 1/F

Explanation:

dC/dt) = (-F/V)*C+(I/V)

To make it easier to solve, let

Constants: I, V,F

Variables: C, t

workings and solution can be viewed below

8 0
3 years ago
Which quantity is proportional to 12/15?
Iteru [2.4K]

4/5

Find common fractions of 12/15

12 / 15. each part divided by 3

15 / 3 = 5...12 / 3 = 4...12/15 = 4/5

6 0
3 years ago
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