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tatuchka [14]
3 years ago
10

What is the mode of 5,6,7,8,9,10

Mathematics
2 answers:
seropon [69]3 years ago
8 0

Answer:

No mode

Step-by-step explanation:

The mode of a data set is the number that occurs the most frequently.

5,6,7,8,9,10

In this data set, each number shows up once, so there is no mode.

Hope this helps! :)

Eduardwww [97]3 years ago
8 0
There is no repeating number or most frequent number meaning there would be no mode in this set of numbers .
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At Sunshine Lanes, each bowler pays $7 per game including shoe rental. Martin graphs the relationship between the number of game
Anna71 [15]

Answer:

The 1st answer that is (0,7) will be No

The 2nd answer that is (3,21) will be Yes

The 3rd answer that is (8,56) will bw Yes

4 0
2 years ago
1. A hotel manager buys 30 pillow cases and 25 shower curtains for $375. The manager then buys another 28 pillow cases and 35 sh
skad [1K]

Answer:

For #2: It's C. 17 nickels and 15 quarters.

Step-by-step explanation:

17*5= 85

15*25= 375

375+85= 460

add in the decimal:

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7 0
3 years ago
SO<br> Determine the values of<br> a, b, and c for<br> the quadratic equation:<br> 4x2 - 8x = 3
Schach [20]
I believe it’s

a = 4
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6 0
3 years ago
The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s
oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is A(6) = 28.

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable t.

Notice that the interval of interest 0 \le t \le 6 is closed on both ends. In other words, this interval includes both endpoints: t = 0 and t= 6. Over this interval, the value of A(t) might be maximized when t is at the following:

  • One of the two endpoints of this interval, where t = 0 or t = 6.
  • A local maximum of A(t), where A^\prime(t) = 0 (first derivative of A(t)\! is zero) and A^{\prime\prime}(t) (second derivative of \! A(t) is smaller than zero.)

Start by calculating the value of A(t) at the two endpoints:

  • A(0) = 10.
  • A(6) = 28.

Apply the power rule to find the first and second derivatives of A(t):

\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}.

\displaystyle A^{\prime\prime}(t) = 6\, t - 15.

Notice that both t = 1 and t = 4 are first derivatives of A^{\prime}(t) over the interval 0 \le t \le 6.

However, among these two zeros, only t = 1\! ensures that the second derivative A^{\prime\prime}(t) is smaller than zero (that is: A^{\prime\prime}(1) < 0.) If the second derivative A^{\prime\prime}(t)\! is non-negative, that zero of A^{\prime}(t) would either be an inflection point (ifA^{\prime\prime}(t) = 0) or a local minimum (if A^{\prime\prime}(t) > 0.)

Therefore \! t = 1 would be the only local maximum over the interval 0 \le t \le 6\!.

Calculate the value of A(t) at this local maximum:

  • A(1) = 15.5.

Compare these three possible maximum values of A(t) over the interval 0 \le t \le 6. Apparently, t = 6 would maximize the value of A(t)\!. That is: A(6) = 28 gives the maximum value of \! A(t) over the interval 0 \le t \le 6\!.

However, note that the maximum over this interval exists because t = 6\! is indeed part of the 0 \le t \le 6 interval. For example, the same A(t) would have no maximum over the interval 0 \le t < 6 (which does not include t = 6.)

4 0
3 years ago
Decrease 1800 in the ratio 5:3​
oksano4ka [1.4K]

Answer:

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Step-by-step explanation:

a decrease in the ratio 5:3 implies that:

new quantity :old quantity :5:3

let new quantity be x.

x:1800=5:3

x/1800=5/3

3x=9000

x=9000/3

x=3000

8 0
2 years ago
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