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3 years ago

Does this set of ordered pairs represent a function? (-1,5)(0,-3)(2,7)(4,0)(7,5)

Mathematics

2 answers:

Lelechka [254]3 years ago

8 0

Answer:

it is indeed a function

Step-by-step explanation:

ruslelena [56]3 years ago

6 0

Answer:

Step-by-step explanation:

You cant have 2 y values

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Georgia out found that 2 out of 25 apples had bruiseson them. based on this information, how many apples out of 100 would be exp

TEA [102]

We just need to extrapolate the information we have about 25 apples to 100, that is we need to find a number that multiplied by gives us 100, from this original fraction, hence, we have:
2 out of 25 = 2/25
if we multiply and divide a number or fraction by the same number we do not alter its value, so we need to convert the 25 into 100, and that can be done multiplying it by 4, like so:
(2/25)(4/4) = (2*4)/(25*4) = 8/100

So if we have 2 bruised apples out of 25, then we can expect to have 8 bruised apples out of 100

6 0

3 years ago

What is the quotient in simplified form? State any restrictions on the variable. Show Work.

photoshop1234 [79]

a+2 a +1
= ------- / ------------------
a -5 (a - 3)(a - 5)

a+2 (a - 3)(a - 5)
= ------- * ------------------
a -5 a +1

(a+2)(a - 3)
= ----------------
a +1

a NOT equal 5 and -1

5 0

3 years ago

Read 2 more answers

Find the arc length of the partial circle.

Step2247 [10]

Arc length of the quarter circle is 1.57 units.
Solution:
Radius of the quarter circle = 1
Center angle (θ) = 90•

Arc length = 1.57 units
Arc length of the quarter circle is 1.57 units.

3 0

3 years ago

A private and a public university are located in the same city. For the private university, 1038 alumni were surveyed and 647 sa

Snezhnost [94]

Answer:

The difference in the sample proportions is not statistically significant at 0.05 significance level.

Step-by-step explanation:

Significance level is missing, it is α=0.05

Let p(public) be the proportion of alumni of the public university who attended at least one class reunion

p(private) be the proportion of alumni of the private university who attended at least one class reunion

Hypotheses are:

$H_{0}$ : p(public) = p(private)

$H_{a}$ : p(public) ≠ p(private)

The formula for the test statistic is given as:

z= $\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of public university students who attended at least one class reunion ( $\frac{808}{1311}=0.616$ )

p2 is the sample proportion of private university students who attended at least one class reunion ( $\frac{647}{1038}=0.623$ )

p is the pool proportion of p1 and p2 ( $\frac{808+647}{1311+1038}=0.619$ )

n1 is the sample size of the alumni from public university (1311)

n2 is the sample size of the students from private university (1038)

Then z= $\frac{0.616-0.623}{\sqrt{{0.619*0.381*(\frac{1}{1311} +\frac{1}{1038}) }}}$ =-0.207

Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.

6 0

3 years ago

Brianna is going to a carnival that has games and rides. Each game costs $1.25 and each ride costs $3.75. Brianna spent $18.75 a

maxonik [38]

Answer:

6 games and 3 rides

Step-by-step explanation:

Let the number of games Brianna played be x,

and the number of rides she went on be y.

Total cost for games= 1.25x

Total cost for rides= 3.75y

Since number of games is twice the number of rides,

x= 2y -----(1)

Total costs= cost of games +cost of rides

$18.75= 1.25x +3.75y$

Multiply the whole equation by 4 to remove the decimals:

$75= 5x +15y$

Simplify by dividing the whole equation by 5:

$15 = x + 3y$

Label the equation:

x +3y= 15 -----(2)

Although we can solve these 2 equations by substitution, since the question requires us to graphically solve, we have to graph 2 linear lines.

I will choose 3 points to plot on the graph for each equation:

x= 2y -----(1)

$\begin{tabular}{|c|c|c|c|} \cline{1-4}x & 2(1) = 2&2(2) = 4&2(3) = 6\\ \cline{1-4}y & 1 &2&3\\ \cline{1-4}\end{tabular}$

x +3y= 15 -----(2)

x= -3y +15

$\begin{tabular}{|c|c|c|c|} \cline{1-4}x & -3(1)+15= 12& -3(2)+15= 9& -3(3)+15= 6\\ \cline{1-4}y & 1 &2&3\\ \cline{1-4}\end{tabular}$

Let's plot these points on a graph paper. Then, join them with a straight line for each straight line graph. Please see the attached picture for the graph.

From (1): y= ½x

From (2): 3y= 15 -x

y= 5 -⅓x

From the graph, the solution of the equation is (6,3). The solution is the point on the graph in which the 2 lines intersect.

x- coordinate: 6

y- coordinate: 3

Thus, Brianna played 6 games and went on 3 rides.

8 0

4 years ago

Read 2 more answers