Find a linear function that models the data in the table. f(x)= *blank* x + *blank*

Student Council sponsors weekly dances at their school on Friday nights. The admission price for each person is $4 for Student C

Morgarella [4.7K]

Answer:

$110

Step-by-step explanation:

So first you have to do 15*4. We get the 4 from the admission fees and the 15 from the number of dances they attend. From this answer we can see the total of money they will be paying for admission for the 15 dances.

15*4=$60

However, our answer isn't just $50 as we are also told membership costs $50. So no we add $50 to $60, which gives us an answer of $110, which is the amount the member will pay if they attend 15 dances during the school year.

8 0

2 years ago

Solve for x •26.45 •10.93 •14.26 •22.19 •20.26

ollegr [7]

Answer:

x = 26.45

Step-by-step explanation:

Cos theta = adjacent side / hypotenuse

cos 50 = 17 /x

Switch the x and the cos 50

x = 17 / cos 50

x =26.44730506

Rounding to 2 decimal places

x = 26.45

8 0

3 years ago

(4y+3)-(2y-9) Simplify. Find the difference.

Nadusha1986 [10]

Answer:

2y+12

Hope this helps

Step-by-step explanation:

5 0

3 years ago

The water diet requires you to drink two cups of water every half hour from the time you get up until you go to bed, but otherwi

Elan Coil [88]

Answer:

$7-3.182 \frac{13.638}{\sqrt{4}}= -14.698$

$7+3.182 \frac{13.638}{\sqrt{4}}= 28.698$

Step-by-step explanation:

For this case we have the following info given:

Weight before diet 180 125 240 150

Weight after diet 170 130 215 152

We define the random variable $D = before-after$ and we can calculate the inidividual values:

D: 10, -5, 25, -2

And we can calculate the mean with this formula:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

And the deviation with:

$s= \sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

And after replace we got:

$\bar D= 7, s_d = 13.638$

And the confidence interval for this case would be given by:

$\bar D \pm t_{\alpha/2} \frac{s_d}{\sqrt{n}}$

The degrees of freedom are given by:

$df = n-1= 4-1=3$

For the 95% of confidence the value for the significance is $\alpha=0.05$ and the critical value would be $t_{\alpha/2}= 3.182$ . And replacing we got: