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3 years ago

the mode (rounded to the nearest tenth if needed) of the following data set is 30,29,40,51,39,33,24,60,40,35,58

2 answers:

7 0

The mode is the number that occurs the most in the sequence of numbers, so the mode is 40 since that number is repeated the most in the sequence, 2 times.

grandymaker [24]3 years ago

5 0

The mode is the number that appears the most.
I'll set them up from least to greatest and in separate rows.
24
29
30
33
35
39
40, 40
51
58
60
The number that appeared the most is 40, so your mode is 40.

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What is the solution to the system of equations?

sukhopar [10]

Answer:

(x, y) = (-3, -6)

Step-by-step explanation:

The second equation can be used to write an expression for y:

... -5x -21 = y . . . . . . . add y-21 to both sides

This expression can be substituted for y in the first equation:

... 6 = -4x +(-5x -21)

... 27 = -9x . . . . . add 21, collect terms

... -3 = x . . . . . . . divide by -9

Using this value of x in the expression for y, we find ...

... -5(-3) -21 = y = -6

The solution is x = -3, y = -6.

8 0

3 years ago

The history museum is hosting a special exhibition on history of flight. The admission ticket for each adult costs $7 and the ad

stiks02 [169]

Answer:

Use simultaneous equation for this problem

y= number of adults

x = number of children

3y + 2x = 160

y + x = 60

then we double the second equation

3y + 2x = 160

2y + 2x = 120

we cancel x by elimination

3y - 2y = 160 - 120

y = 40

Step-by-step explanation:

hope this helps

5 0

2 years ago

Solve for a.<br> a^2=-3a+88

Sergio [31]

Answer:

$a=8,-11$

Step-by-step explanation:

$a^2=-3a+88$

$a^2+3a=88$

$a^2+3a-88=0$

$x=\frac{-b\pm\sqrt{b^2-4ac} }{2a}$

$x=\frac{-3\pm\sqrt{3^2-4(1)(-88)} }{2(1)}$

$x=\frac{-3\pm\sqrt{9+352} }{2}$

$x=\frac{-3\pm\sqrt{361} }{2}$

$x=\frac{-3\pm19 }{2}$

$x=-11$ and $x=8$

Therefore, the values of $a$ are 8 and -11.

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2 years ago

Idently the graph of f(x) = 3(x - 1) - 4. Then identity the vertex and ads of symmetry find the minimum value of and describe wh

Vesna [10]

Answer:

See below

Step-by-step explanation:

I assume you mean $f(x) = 3(x-1)^2-4$

The equation is already in vertex form $f(x)=a(x-h)^2+k$ where $a$ affects how "fat" or "skinny" the parabola is and $(h,k)$ is the vertex. Therefore, the vertex is $(h,k)\rightarrow(1,-4)$ .

The axis of symmetry is a line where the parabola is cut into two congruent halves. This is defined as $x=h$ for a parabola with a vertical axis. Hence, the axis of symmetry is $x=1$ .

The minimum value is the smallest value in the range of the function. In the case of a parabola, the y-coordinate of the vertex is the minimum value. Therefore, the minimum value is $y=-4$ .