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3 years ago

12.

1 answer:

8 0

Last year, the average math SAT score for students at one school was 475. The headmaster then introduced a new teaching method hoping to improve scores. This year, the mean math SAT score for a sample of students was 491. The headmaster concluded that the new teaching method produces higher SAT scores. The problem with reporting results this way is voluntary response. The information of how the teaching method isnot mentioned.

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2m+5=5(m-7)-3m equals??

Svetradugi [14.3K]

2m+5=5(m-7)-3m
first, use distributive property...
2m+5=5m-35-3m
next, combine like terms on the right side...
2m+5=2m-35
next, subtract 2m from both sides.
2m+5=2m-35
-2m -2m
you will get....
5=-35 in which is not true. This means the equation has no solution.

6 0

3 years ago

Read 2 more answers

A warehouse has 1,750 boxes of water bottles. A truck unloaded another 530 boxes at the warehouse. Each box holds 48 bottles of

Lapatulllka [165]

Answer:

109440

Step-by-step explanation:

1750 boxes of water bottles with 48 bottles of water each: 1750 x 48 = 84000

plus the other 530 boxes at the warehouse which alone are: 530 x 48 = 25440

adding up 84000 + 25440 = 109440

8 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$