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Svetradugi [14.3K]

3 years ago

5

Six friends went to a baseball game.The price of the admission per person was $x.Three of the friends bought a hot dog for $3. W

rite and simplify an expression that represents the total cost.

1 answer:

Ronch [10]3 years ago

8 0

C(cost)=6x+9
9 is the cost of three hot dogs
6x represents 6 tickets and their cost

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write the standard form of the line that passes through the given points include your work in your final answer type your answer

Gemiola [76]

The slope is change in "Y" divided by the change in "x"
1-(-3)/2-(-1)
= 4/3

Standard Form
y-y1=m(x-x1)
y-(-3)=4/3(x-(-1)
y+3=4/3x+4/3
y=4/3x-5/3

Rewrite in standard form
-4/3x+y=-5/3
Can't have fractions so multiply by 3 to get rid of the denominator
-4x+3y=5
Divide by -1 to get rid of the negative sign

4x-3y=-5

4 0

3 years ago

Find the distance between each pair of parallel line<br>y=x-6 and y=x+8

myrzilka [38]

Both these lines are linear and have the same slope, just different y-intercepts. This means one line is shifted higher than the other. y=x+8 is 14 units higher than y=x-6. The answer to you question therefore is 14 units!

5 0

3 years ago

A journeyperson earns $28.75 per hour. A new apprentice earns 60% of the journeyperson's rate.

maxonik [38]

Answer:

$503.70

Step-by-step explanation:

$28.75 × .6 = $17.25

$17.25 × 40 = $690

$690 × .27 = $186.30

$690 - $186.30 = $503.70

8 0

2 years ago

SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is inter

Rama09 [41]

Answer:

The administrator should sample 968 students.

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.88}{2} = 0.06$

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 p-value of $1 - 0.06 = 0.94$ , so Z = 1.555.

Now, find the margin of error M as such

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

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

Standard deviation of 300.

This means that $n = 300$

If the administrator would like to limit the margin of error of the 88% confidence interval to 15 points, how many students should the administrator sample?

This is n for which M = 15. So

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

$15 = 1.555\frac{300}{\sqrt{n}}$

$15\sqrt{n} = 300*1.555$

Dividing both sides by 15

$\sqrt{n} = 20*1.555$

$(\sqrt{n})^2 = (20*1.555)^2$

$n = 967.2$

Rounding up:

The administrator should sample 968 students.

5 0

3 years ago

A rectangle has a width of 9 units and a length of 40 units. What is the length of a diagonal?

____ [38]

The answer is c. 41 units.

3 0

3 years ago