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3 years ago

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Michael has a home-based business putting on children's parties. He charges $45 to design the party and then $7.50 per child . W

rite a function rule ( equation) that relates the total cost of the party to the number of children n

2 answers:

love history [14]3 years ago

6 0

Answer:

45 + ( 7.50*x)

Step-by-step explanation:

Anika [276]3 years ago

5 0

Answer:

I believe the answer is: $f(x)=45+7.50x$

Step-by-step explanation:

45 because you start with it costing that much and then the 7.50x because each child costs 7.50 and you don't know how many there will be.

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Step-by-step explanation:

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3 years ago

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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

-8+4u=-28<br> Simplify your answer as much as possible

yulyashka [42]

The answer would be -5

6 0

3 years ago

Read 2 more answers

1. Let f(x, y) be a differentiable function in the variables x and y. Let r and θ the polar coordinates,and set g(r, θ) = f(r co

Olenka [21]

Answer:

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}\\$

Step-by-step explanation:

First, notice that:

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}cos(\frac{\pi}{4}),\sqrt{2}sin(\frac{\pi}{4}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(\sqrt{2}(\frac{1}{\sqrt{2}}),\sqrt{2}(\frac{1}{\sqrt{2}}))\\$

$g(\sqrt{2},\frac{\pi}{4})=f(1,1)\\$

We proceed to use the chain rule to find $g_{r}(\sqrt{2},\frac{\pi}{4})$ using the fact that $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ to find their derivatives:

$g_{r}(r,\theta)=f_{r}(rcos(\theta),rsin(\theta))=f_{x}( rcos(\theta),rsin(\theta))\frac{\delta x}{\delta r}(r,\theta)+f_{y}(rcos(\theta),rsin(\theta))\frac{\delta y}{\delta r}(r,\theta)\\$

Because we know $X(r,\theta)=rcos(\theta)\ and\ Y(r,\theta)=rsin(\theta)$ then:

$\frac{\delta x}{\delta r}=cos(\theta)\ and\ \frac{\delta y}{\delta r}=sin(\theta)$

We substitute in what we had:

$g_{r}(r,\theta)=f_{x}( rcos(\theta),rsin(\theta))cos(\theta)+f_{y}(rcos(\theta),rsin(\theta))sin(\theta)$

Now we put in the values $r=\sqrt{2}\ and\ \theta=\frac{\pi}{4}$ in the formula:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=f_{x}(1,1)cos(\frac{\pi}{4})+f_{y}(1,1)sin(\frac{\pi}{4})$

Because of what we supposed:

$g_{r}(\sqrt{2},\frac{\pi}{4})=f_{r}(1,1)=-2cos(\frac{\pi}{4})+3sin(\frac{\pi}{4})$

And we operate to discover that:

$g_{r}(\sqrt{2},\frac{\pi}{4})=-2\frac{\sqrt{2}}{2}+3\frac{\sqrt{2}}{2}$

$g_{r}(\sqrt{2},\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

and this will be our answer

3 0

3 years ago