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3 years ago

The scatter plot below shows the ages and heights of a baseball team. Each dot represents one player. What is the total

Mathematics

2 answers:

Jobisdone [24]3 years ago

8 0

Answer:

its 2 because it says how many are 12 years old and are taller than 60

vlada-n [284]3 years ago

7 0

Answer:

Step-by-step explanation:

if each dot is one person then there are only 2 people that are 12 and that are more than 60in tall

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(-4-3)(1-(3+5))x5= <br><br><br> how do you solve that problem ?

zavuch27 [327]

(-4 - 3)(1 - (3 + 5)) × 5
(-7)(-1 - 2) × 5
(-7)(-3) × 5
21 × 5
105

7 0

3 years ago

–2x + 2y = -4<br> 3х + Зу = -18

rodikova [14]

Answer:

(-2, -4)

Step-by-step explanation:

–2x + 2y = -4

3х + Зу = -18

Divide the first equation by 2

–2x/2 + 2y/2 = -4/2

-x +y = -2

Divide the second equation by 3

3х/3 + Зу/3 = -18/3

x + y = -6

Add these equation together

-x +y = -2

x + y = -6

-------------------

0x + 2y = -8

2y = -8

Divide each side by 2

2y/2 = -8/2

y = -4

Now we need x

x+y = -6

x-4 = -6

Add 4 to each side

x-4+4 = -6+4

x = -2

3 0

3 years ago

The table shows the concentration of a reactant in the reaction mixture over a period of time.

saveliy_v [14]

<h3>Option C</h3><h3>The average rate of the reaction over the entire course of the reaction is: $2.0 \times 10^{-3}$ </h3>

<em><u>Solution:</u></em>

Average rate is the ratio of concentration change to the time taken for the change

$Average\ rate = \frac{ \triangle c }{\triangle t }$

The concentration of the reactants changes 1.8 M to 0.6 M

here, the time interval given is 0 to 580 sec

Therefore,

$Average\ rate = \frac{1.8-0.6}{580} \\\\Average\ rate = \frac{1.2}{580} \\\\Average\ rate \approx 2.0 \times 10^{-3}$

Thus option C is correct

5 0

3 years ago

Describe the location of the point having the following coordinates.

denis23 [38]

we know that

If the point has an nonzero abscissa (x coordinate) then it can be located in any of the four quadrants

If the point has an negative ordinate (y coordinate) then it can be located in Quadrant III or Quadrant IV

therefore

the answer is the option

Quadrant III or Quadrant IV

7 0

3 years ago

The height after t seconds of an object projected upward with an initial velocity of 48 feet per second from a 210-foot tower ca

svetlana [45]

Given:

The height h of an object after t seconds is

$h=-16t^2+48t+210$

The height of a neighboring 50-foot tall building is modeled by the equation h=50.

The time (t) when the object will be at the same height as the building is found to be t = –2 and t = 5.

To find:

The statement which describes the validity of these solutions.

Solution:

We have,

$h=-16t^2+48t+210$

Here, t is the time in seconds.

For t=-2,

$h=-16(2)^2+48(-2)+210$

$h=-64-96+210$

$h=50$

For t=5,

$h=-16(5)^2+48(5)+210$

$h=-400+240+210$

$h=50$

So, the value of h is 50 at t=-2 and t=5.

We know that time is always positive so it cannot be negative value. It means t=-2 is not possible.

The solution t = 5 is the only valid solution to this system since time cannot be negative.

Therefore, the correct option is C.

6 0

3 years ago