Ask question

Ask question

All categories

2 years ago

9

Daisy purchases a gym membership.She pays a signup fee and a monthly fee off 13$.After 4 months, she has paid a total of$69.Use

a linear equation in point-slope form to find the signup fee.Solve the problem using an equation in point-slope form.

1 answer:

bulgar [2K]2 years ago

4 0

Answer:

The signup fee is $17

Step-by-step explanation:

we know that

The equation in point slope form is equal to

$y-y1=m(x-x1)$

where

m is the slope or unit rate of the linear equation

(x1,y1) is a ordered pair of the line

Let

x -----> the number of months

y -----> the total cost of a gym membership

In this problem we have

The slope of unit rate is equal to

$m=\$13\ per\ month$

The ordered pair is

$(x1,y1)=(4,69)$

substitute

$y-69=13(x-4)$ ----> equation in point slope form

Convert to slope intercept form

$y-69=13x-52$

$y=13x-52+69$

$y=13x+17$ ---> equation in slope intercept form

The y-intercept is the signup fee (value of y when the value of x is equal to zero)

therefore

The signup fee is $17

You might be interested in

How to define when something is certain to happen?

PIT_PIT [208]

You could say invitability or a foregone conclusion. Invitability just means "the quality of being certian to happen."

Example: "My teacher said it was invitable that she was going to give us a math test tomorrow"

3 0

3 years ago

Which expression is a trinomial?<br> 2x² + 2x - 4<br><br> 7x^3 - 2X<br><br> 50x<br><br> 2x²

andreev551 [17]

The first option: 2x^2+2x-4 since there are three terms

5 0

2 years ago

Read 2 more answers

Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

2 years ago

- Write a system of equations in which one equation could be multiplied by-2 and the

iris [78.8K]

2x+5y=9

3x+7y=13

6x+15y=27

-6x-14y=-26

y=1

x=2

6 0

2 years ago

Which dimensions can create only one unique triangle?

s2008m [1.1K]

Answer:

<h2>The dimensions must all add up to 180°.</h2>

Step-by-step explanation:

For example;

Your triangles' angles are: 56, 77, and 47.

56 + 77 + 47 = 180

*You did not give me any values to solve for, but just since you didn't- I'v provided an example to assist you in doing it yourself. If you want me to solve your full equation please add the angles or 3 dimensions of the triangle.

<h3><em>I hope this helped!</em></h3>

3 0

2 years ago