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sdas [7]
3 years ago
6

Please help! Correct answer only!

Mathematics
1 answer:
bonufazy [111]3 years ago
8 0

Answer:

<em>Expected Payoff ⇒ $ 1.50 ; Type in 1.50</em>

Step-by-step explanation:

Considering that 1 out of the 100 tickets will have a probability of winning a 150 dollar prize, take a proportionality into account;

100 - Number of Tickets,\\1 - Number of Tickets You Can Enter,\\\\1 / 100 - Probability of Winning,\\$ 150 - Money Won,\\\\Proportionality - 1 / 100 = x / 150, where x - " Expected Payoff "\\\\1 / 100 = x / 150,\\100 * x = 150,\\\\Conclusion ; x = 1.5 dollars

<em>Thus, Solution ; Expected Payoff ⇒ $ 1.50</em>

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3-(-4) answer the question
Jobisdone [24]

Answer:

<h2>7</h2>

Step-by-step explanation:

3-(-4) \\-\times - = +\\3+4 \\=7

7 0
4 years ago
Read 2 more answers
A full lake begins dropping at a constant rate. After 4 weeks it has dropped 3 feet. What is the unit rate of change in the lake
Valentin [98]
<span>A full lake is dropping at its constant rate, meaning it is unchanging over time.
After 4 weeks’ time, it already dropped 3 feet.
Now, let’s solve for the Unit rate based on the given data.
=> 3 feet in 4 weeks
=> Unit rate is 3 feet per 4 weeks
=> in 4 weeks there are (7 * 4) = 28 days
=> 3 feet / 28 days  
=> the Unit rate is .107 feet / day</span>



7 0
3 years ago
How many gallons and quarts are in 10 quarts
Ipatiy [6.2K]
10 quarts = 2.5 gallons.

There are 4 quarts for every 1 gallon, but because you're finding gallons and not quarts, you do 10 ÷ 4 = 2.5.
7 0
4 years ago
Read 2 more answers
The fish population in a pond with carrying capacity 1200 is modeled by the following logistic equation where N(t) denotes the n
MArishka [77]

9514 1404 393

Answer:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

  142 fish

Step-by-step explanation:

A) The differential equation is modified by adding a -50 fish per year constant term:

  (dN)/(dt) = (0.4)/(1200)N(1200-N) -50

__

B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.

  (0.4/1200)(N)(1200-N) -50 = 0

  N(N-1200) = -(50)(1200)/0.4) . . . . rewrite so N^2 has a positive coefficient

  N^2 -1200N + 600^2 = -150,000 +600^2 . . . . complete the square

  (N -600)^2 = 210,000 . . . . . simplify

  N = 600 + √210,000 ≈ 1058

This steady-state number of fish is ...

  1200 - 1058 = 142 . . . . below the original carrying capacity

8 0
3 years ago
(6 points) Elderly drivers. A polling agency interviews 748 American adults and finds that 464 think licensed drivers should be
NNADVOKAT [17]

Answer:

1) p-hat = 0.6203

2) std error = 0.0177

3) margin of error = 0.0291

4) (0.5912, 0.6494)

5) minimum sample size = 1683

Step-by-step explanation:

1.  The point estimate for the proportion of American adults that think licensed drivers should be required to retake their road test once they reach 65 years of age can be calculating dividing the number of people who answer Yes by the total of people interviewed:

\hat p=X/N=464/748=0.6203

2. The standard error of a proportion can be expressed as:

se_p=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.6203*0.3797}{748}}=\sqrt{0.000314877}=0.0177

3. The z-value for a 90% CI is z=1.64485.

The margin of error can be expressed as:

me=z\cdot se_p = 1.64485*0.0177=0.0291

4. The lower and upper limits can be calculated as:

LL=\hat p - z*se_p=0.6203-0.0291=0.5912\\\\\\UL=\hat p + z*se_p=0.6203+0.0291=0.6494

5. To have a margin of error of no more than 3.2% of the sample proportion, we have to first calculate what margin of error gives 3.2% of the proportion.

me=0.032*\hat p=0.032*0.6203=0.0198

This margin of error can be related to the sample size as:

me=z*se_p=z*\sqrt{\frac{p(1-p)}{n}} =1.645\sqrt{\frac{0.6203*0.3797}{n}}=0.0198\\\\1.645\sqrt{\frac{0.2355}{n} }=0.0198\\\\  \frac{0.2355}{n} =(0.0198/1.645)^2=0.0120^2=0.00014\\\\n=0.2355/0.00014\\\\n=1682.14\approx1683

5 0
3 years ago
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