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Nonamiya [84]
3 years ago
10

which table represents viable solutions for y=5x, where x is the number of tickets sold for the school play a y is the amount of

money collected for the tickets?

Mathematics
1 answer:
STALIN [3.7K]3 years ago
8 0

Given:

The given equation is:

y=5x

Where x is the number of tickets sold for the school play a y is the amount of money collected for the tickets.

To find:

The correct table of values from the given options.

Solution:

We know that the number of tickets and amount of money collected for the tickets cannot be negative. So, options A and B are incorrect.

We have,

y=5x

For x=0,

y=5(0)

y=0

For x=10,

y=5(10)

y=50

For x=51,

y=5(51)

y=255

For x=400,

y=5(400)

y=2000

In option C, all the ordered pairs (0,0), (10,50), (51,255), (400,2000) are in the table. So, option C is correct.

For x=65

y=5(65)

y=325

Since the ordered pair (65,350) does not satisfy the given equation, therefore the option D is incorrect.

Hence, the correct option is C.

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The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp
andreyandreev [35.5K]

Answer:

(a) X ~ N(\mu=63, \sigma^{2} = 13^{2}).

    \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = <u><em>amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the normal distribution is given by;

                      Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

So, the distribution of X ~ N(\mu=63, \sigma^{2} = 13^{2}).

Let \bar X = <u><em>sample mean amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the sample mean is given by;

                      Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

            n = sample of people = 43

So, the distribution of \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

   P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X \leq 61.4 mL)

  P(X < 62.8 mL) = P( \frac{X-\mu}{\sigma} < \frac{62.8-63}{13} ) = P(Z < -0.02) = 1 - P(Z \leq 0.02)

                                                           = 1 - 0.50798 = 0.49202

  P(X \leq 61.4 mL) = P( \frac{X-\mu}{\sigma} \leq \frac{61.4-63}{13} ) = P(Z \leq -0.12) = 1 - P(Z < 0.12)

                                                           = 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < \bar X < 62.8 mL)

   P(61.4 mL < \bar X < 62.8 mL) = P(\bar X < 62.8 mL) - P(\bar X \leq 61.4 mL)

  P(\bar X < 62.8 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{62.8-63}{\frac{13}{\sqrt{43} } } ) = P(Z < -0.10) = 1 - P(Z \leq 0.10)

                                                           = 1 - 0.53983 = 0.46017

  P(\bar X \leq 61.4 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{61.4-63}{\frac{13}{\sqrt{43} } } ) = P(Z \leq -0.81) = 1 - P(Z < 0.81)

                                                           = 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

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Semmy [17]
The answer is 440.
What I do is just take the -11 and -15 out of the parentheses so it is -11*-15 and the answer is 165. Then, you take -11 and -25 out of the parentheses and find the answer.
Add them together.
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Answer:

x\geq 6\text{ or } x

Step-by-step explanation:

We have the compound inequality:

2(x-1)\geq10\text{ or } 3-4x>15

Let's solve each of them individually first:

We have:

2(x-1)\geq10

Divide both sides by 2:

x-1\geq5

Add 1 to both sides:

x\geq6

We have:

3-4x>15

Subtract from both sides:

-4x>12

Divide both sides by -4:  

x

Hence, our solution set is:

x\geq 6\text{ or } x

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