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3 years ago

The quantities x and y are proportional.

Mathematics

1 answer:

Ksenya-84 [330]3 years ago

4 0

Answer:

r=3,r=25/27,r=3

Step-by-step explanation:

In equation y=rx

make r the subject of the formula by dividing both sides by x

r=y/x

1). when y=15 and x=5

r=15/5

r=3

2). when y=25 and x=81/3

r=25/81/3

r=25÷81/3

r=25*3/81

r=75/27

r=25/27

3). when y=33 and x=11

r=33/11

r=3

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450 high school freshmen were randomly selected for a national survey. Among survey participants, the mean grade-point average (

umka21 [38]

Answer:

The margin of error is of 0.01.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.7}{2} = 0.15$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.15 = 0.85$ , so Z = 1.037.

The margin of error is of:

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation was 0.21.

This means that $\sigma = 0.21$

Sample of 450:

This means that $n = 450$

What is the margin of error, assuming a 70% confidence level, to the nearest hundredth?

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 1.037\frac{0.21}{\sqrt{450}}$

$M = 0.01$

The margin of error is of 0.01.

4 0

3 years ago

17.

777dan777 [17]

Answer:

$12.5

Step-by-step explanation:

Since prize is split equally.

So, money received by each person

= 75/6

= $12.5

7 0

3 years ago

In mrs. wiggins class of 32 students, 3/8 of the student are taking French, and 1/2 of the students are taking Latin. If 6 stude

Vlad1618 [11]

The answer is 10 you see 6 people are in both of the classes so you have to subtract 6 from the 12 and 16 12+16-6=22
32-22=10

5 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:

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