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Anna71 [15]
3 years ago
15

Fiona needs $62 to go on a field trip. She has saved $10.50. She earns $6.50 per hour cleaning her neighbor's garden, and she ea

rns $5.25 per hour selling cakes at a store. Can Fiona go on the field trip if she works at the garden for 6 hours and at the store for 2 hours?
(Use the inequality 6.50x + 5.25y + 10.50 ≥ 62.)
Mathematics
1 answer:
3241004551 [841]3 years ago
3 0

Answer:

Fiona cannot go on the field trip

Step-by-step explanation:

6.5 x 6 = 49

5.25 x 2 = 10.5

49 + 10.5 + 10.5 = 60

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I dont really understand this??​
jeyben [28]

Answer:

x\geq  -53

Step-by-step explanation:

The inequality includes -53.

The interval notation is \:[-53,\:\infty \:)

\{ x \in \mathbb{R}: x \geq  -53 \}

8 0
3 years ago
Read 2 more answers
The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponenti
melomori [17]

Answer:

a) 0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

b) Capacity of 252.6 cubic feet per second

Step-by-step explanation:

Exponential distribution:

The exponential probability distribution, with mean m, is described by the following equation:

f(x) = \mu e^{-\mu x}

In which \mu = \frac{1}{m} is the decay parameter.

The probability that x is lower or equal to a is given by:

P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

Which has the following solution:

P(X \leq x) = 1 - e^{-\mu x}

The probability of finding a value higher than x is:

P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second).

This means that m = 100, \mu = \frac{1}{100} = 0.01

(a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

We have that:

P(X > x) = e^{-\mu x}

This is P(X > 190). So

P(X > 190) = e^{-0.01*190} = 0.1496

0.1496 = 14.96% probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day.

(b) What water-pumping capacity, in cubic feet per second, should the station maintain during early afternoons so that the probability that demand will exceed capacity on a randomly selected day is only 0.08?

This is x for which:

P(X > x) = 0.08

So

e^{-0.01x} = 0.08

\ln{e^{-0.01x}} = \ln{0.08}

-0.01x = \ln{0.08}

x = -\frac{\ln{0.08}}{0.01}

x = 252.6

Capacity of 252.6 cubic feet per second

5 0
3 years ago
Can someone please help me?
Vadim26 [7]

Answer:

0.84

Step-by-step explanation:

hope this helps, brainliest appreciated!

8 0
2 years ago
In a volcano, erupting lava flows continuously through a tube system about 20 kilometers to the sea. Assume a lava flow of 0.5 k
butalik [34]

Answer:

40 hours

Step-by-step explanation:

Lets take "x" as the # of hours it takes the lava to reach the sea

since there is a lava flow of 0.5 kilos per hour

we will multiply "x" with 0.5

0.5x

now the tube system is 20 kilometers long

this means that at a rate of 0.5 kilos per hour the lava will reach the sea once it is 20 kilometers deep

so

0.5x=20

divide "0.5" by both sides to eliminate the coefficient

\frac{0.5x}{0.5} =\frac{20}{0.5}

x=40

it will take 40 hours for the lava to reach the sea

Hope this helps!

3 0
3 years ago
Question 14. A carousel makes one complete revolution every 75 seconds. Levi sits on the carousel 4 feet from its center. If the
garik1379 [7]

Answer:

The correct option is;

H. 32·π

Step-by-step explanation:

The given information are;

The time duration for one complete revolution = 75 seconds

The distance from the center of the carousel where Levi sits = 4 feet

The time length of a carousel ride = 5 minutes

Therefore, the number of complete revolutions, n, in a carousel ride of 5 minutes is given by n = (The time length of a carousel ride)/(The time duration for one complete revolution)

n = (5 minutes)/(75 seconds) = (5×60 seconds/minute)/(75 seconds)

n = (300 s)/(75 s) = 4  

The number of complete revolutions - 4

The distance of 4 complete turns from where Levi seats = 4 ×circumference of circle of Levi's motion

∴ The distance of 4 complete turns from where Levi seats = 4 × 2 × π × 4 = 32·π.

3 0
3 years ago
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