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cluponka [151]
2 years ago
8

Eddie's fishing. He has 1 1/4 ounces of weights on his line, but his bait isn't getting to the bottom of the lake. Eddie add ano

ther 1/2-ounces weight to the line, but it's still not enough. Finally, Eddie adds a 3/4-ounces weights and the bait sinks. How much weight does eddie now have on his line?
Mathematics
2 answers:
mario62 [17]2 years ago
8 0
The initial weight is given by:
 1  \frac{1}{4}  = 1.25

 Then, we add a second weight:
 \frac{1}{2}  = 0.5

 Finally, we add a third weight:
 \frac{3}{4}  = 0.75

 Adding the three added weights, we can find the total weight.
 We have then:
 1.25 + 0.5 + 0.75 = 2.5

 The total weight is:
 2.5 = 2  \frac{1}{2} 

 Answer:
 
eddie now has 2  \frac{1}{2} ounces on his line
mel-nik [20]2 years ago
8 0

Answer:

idk

Step-by-step explanation:

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Martin and three of his friends take turns driving on a trip. The gas mileage for their car ranges from 28.94 to 24.56 miles per
mart [117]

Answer:

A. No, the estimate should be higher

Step-by-step explanation:

To determine the maximum number of gallons of fuel the car can consume, we express this as;

Maximum miles driven=highest consumption per gallon×Maximum gallons

where;

Maximum miles driven per person=64 mile

Maximum miles driven=64×4=256 miles

highest consumption per gallon=24.56 miles per gallon

Maximum gallons=g

replacing;

256=24.56×g

24.56 g=256

g≥256/24.56=10.42 gallons

g≥10.42 gallons

No, the estimate should be higher

8 0
2 years ago
Air pressure may be represented as a function of height above the surface of the Earth as shown below. P(h)=PoE^-.00012h In this
AlexFokin [52]

The equation to find the air pressure (P(h)) at a height above sea level (h) is given by

P(h)=P_{0} e^{-0.00012h}

Where,

  • P(h) is the new pressure at height h
  • P_{0} is the air pressure at sea level, and
  • h is the height (measured in meters).

We want to know the new pressure that is 65% of P_0.  So we should substitute 0.65P_0 in P(h) (new pressure) to get the equation:

0.65P_0=P_0e^{-0.00012h}. Answer choice B is the correct one.

ANSWER: B

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2 years ago
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suppose you have a part-time job delivering packages. Your employer pays you a flat rate of $9.50 per hour. You discover that th
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must a function that is decreasing over a given interval always be negative over that same interval explain
kolbaska11 [484]

check the picture below.


notice, a function that is decreasing, namely  have a negative slope, doesn't have to be negative.

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2 years ago
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A really bad carton of eggs contains spoiled eggs. An unsuspecting chef picks eggs at random for his ""Mega-Omelet Surprise."" F
Dima020 [189]

Answer:

(a) The probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b) The probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c) The probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

Step-by-step explanation:

The complete question is:

A really bad carton of 18 eggs contains 8 spoiled eggs. An unsuspecting chef picks 5 eggs at random for his “Mega-Omelet Surprise.” Find the probability that the number of unspoiled eggs among the 5 selected is

(a) exactly 5

(b) 2 or fewer

(c) more than 1.

Let <em>X</em> = number of unspoiled eggs in the bad carton of eggs.

Of the 18 eggs in the bad carton of eggs, 8 were spoiled eggs.

The probability of selecting an unspoiled egg is:

P(X)=p=\frac{10}{18}=0.556

A randomly selected egg is unspoiled or not is independent of the others.

It is provided that a chef picks 5 eggs at random.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 5 and <em>p</em> = 0.556.

The success is defined as the selection of an unspoiled egg.

The probability mass function of <em>X</em> is given by:

P(X=x)={5\choose x}(0.556)^{x}(1-0.556)^{5-x};\ x=0,1,2,3...

(a)

Compute the probability that of the 5 eggs selected exactly 5 are unspoiled as follows:

P(X=5)={5\choose 5}(0.556)^{5}(1-0.556)^{5-5}\\=1\times 0.05313\times 1\\=0.0531

Thus, the probability that of the 5 eggs selected exactly 5 are unspoiled is 0.0531.

(b)

Compute the probability that of the 5 eggs selected 2 or less are unspoiled as follows:

P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

              =\sum\imits^{2}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=0.0173+0.1080+0.2706\\=0.3959

Thus, the probability that of the 5 eggs selected 2 or less are unspoiled is 0.3959.

(c)

Compute the probability that of the 5 eggs selected more than 1 are unspoiled as follows:

P (X > 1) = 1 - P (X ≤ 1)

              = 1 - P (X = 0) - P (X = 1)

              =1-\sum\limits^{1}_{x=0}{{5\choose 5}(0.556)^{5}(1-0.556)^{5-5}}\\=1-0.0173-0.1080\\=0.8747

Thus, the probability that of the 5 eggs selected more than 1 are unspoiled is 0.8747.

6 0
2 years ago
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