Please help!

Please someone please help, will mark you Brainliest

Gnoma [55]

Answer:

Step-by-step explanation:

Community gym

=> membership fee = $50

=> monthly fee = $55

=> First month = 105$

=> 50 + 55(x)

=> 50 + 55 (15)

=> 50 + 825

=> 875

Workout gym

=> membership fee = $200

=> monthly fee = $45

=> First month = 245$

=> 200 + 45(x)

=> 200 + 45(15)

=> 200 + 675

=> 875

On the 15th month, community gym and workout gym will both get the same amount of $875.

4 0

3 years ago

Read 2 more answers

If one postage stamp costs $0.85, how much does a book of 25 stamps cost?

aniked [119]

Answer: $21.25

Step-by-step explanation:

0.85 x 25

5 0

3 years ago

Read 2 more answers

How many elements are in the set (a,b, c.?

diamong [38]

There are three elements in that set.

5 0

3 years ago

Find the (a) mean, (b) median, (c) mode, and (d) midrange for the data and then (e) answer the given question. Listed below

mafiozo [28]

Answer:

a) $\bar X = 369.62$

b) $Median=175$

c) $Mode =450$

With a frequency of 4

d) $MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

e) $s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

Step-by-step explanation:

We have the following data set given:

49 70 70 70 75 75 85 95 100 125 150 150 175 184 225 225 275 350 400 450 450 450 450 1500 3000

Part a

The mean can be calculated with this formula:

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

Replacing we got:

$\bar X = 369.62$

Part b

Since the sample size is n =25 we can calculate the median from the dataset ordered on increasing way. And for this case the median would be the value in the 13th position and we got:

$Median=175$

Part c

The mode is the most repeated value in the sample and for this case is:

$Mode =450$

With a frequency of 4

Part d

The midrange for this case is defined as:

$MidR= \frac{Max +Min}{2}= \frac{49+3000}{2}= 1524.5$

Part e

For this case we can calculate the deviation given by:

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

And replacing we got:

$s = 621.76$

And we can find the limits without any outliers using two deviations from the mean and we got:

$\bar X+2\sigma = 369.62 +2*621.76 = 1361$

And for this case we have two values above the upper limit so then we can conclude that 1500 and 3000 are potential outliers for this case

5 0

3 years ago

Explain why the graph of quadratic function could not contain both a minimum vertex and a maximum vertex at the same time.

KonstantinChe [14]

A quadratic function is a function of the form $f(x)=ax^2+bx+c$ . The vertex, $(h,k)$ of a quadratic function is determined by the formula: $h= \frac{-b}{2a}$ and $k=f(h)$ ; where $h$ is the x-coordinate of the vertex and $k$ is the y-coordinate of the vertex. The value of $a$ determines if the parabola opens upward or downward; if $a$ is positive, the parabola opens upward and the vertex is the minimum value, but if $a$ is negative the graph opens downward and the vertex is the maximum value. Since the quadratic function only has one vertex, it could not contain both a minimum vertex and a maximum vertex at the same time.

3 0

3 years ago