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MA_775_DIABLO [31]
3 years ago
13

Which of the following sentences does not represent a function?

Mathematics
2 answers:
kkurt [141]3 years ago
8 0

Answer:

D.

Step-by-step explanation:

A function is represented by every input has only one output. Let’s look at each of the answer choices:

A. Input 4 notebooks, cost is $20 and input 5 notebooks, cost is $25. This is a function

B. Input 1 hour, charge is $20 and input 2 hours, charge is $40. This is a function

C. Input 1 hour, drove 50 miles and input 2 hours, drove 100 miles. This is a function.

D. Input 6 gum balls chewed, chewed in 2 and 3 hours. This is NOT a function. The input does not have only one output.

I hope this helps! :)

umka2103 [35]3 years ago
5 0

Answer:

D

Timon chewed six gumballs in two hours today and six gumballs in three hours yesterday.

Step-by-step explanation:

You might be interested in
NEED HELP ASAP PLEASE
katen-ka-za [31]

Answer:

Pretty sure its 40

Step-by-step explanation:

There is only 3 odd numbers on a 6 sided dice (1, 3, 5) and you would just divide 120 and 3

5 0
4 years ago
The solutions to a certain quadratic equation are x = -4 and x = 3. Write the equation in standard form below.
saw5 [17]

Answer:

x^2 + x -12

Step-by-step explanation:

If the solutions are x=-4 and x=3, then it must have factors (x+4)(x-3). Assuming there is no GCF or leading coefficient other than 1, multiply with FOIL to find the standard form.

(x+4)(x-3)\\x^2 + 4x -3x -12\\x^2 + x -12

7 0
3 years ago
Read 2 more answers
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.

4 0
3 years ago
Find the scale factor of the similar shapes below.
timofeeve [1]

Answer:

x=12 and the scale factor is 1.5

Step-by-step explanation:

From D' to F' is 7.5 and from D to F is 5cm the difference is 2.5. From F' to E' is x and from F to E is 8. This is just understanding what you can see. You find x by figuring out the ratio from 5 to 7.5 is and 8 to x or vise versa.  It depends which way you are going. I think you are going from E,D,F to E',D',F'. So from 5cm to 7.5 cm. This is 1.5 because you are dividing 7.5/5. If you needed to find x, it would be 12.

5 0
3 years ago
Find the area of rectangle it its length is (3x+5) and breadth is (2y+4)
Sindrei [870]

Answer:

Solution given:

length = (3x+5)

breadth = (2y+4)

we have

area of rectangle: length* breadth

=(3x+5)(2y+4)

opening bracket

=3x(2y+4)+5(2y+4)

=6xy+12x+10y+20

=<u>12x+10y+6xy+20</u><u>u</u><u>n</u><u>i</u><u>t</u><u> </u><u>square</u>

4 0
3 years ago
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