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FrozenT [24]
3 years ago
7

We are planning a reception for our daughter. The hall that we are renting charges a $90 cleanup fee. In addition to the cleanup

fee, they charge $29 per guest for food and drinks.
If we have $3,100.00 available, what is the maximum number of guest we can invite without going over our budget?
Mathematics
1 answer:
iris [78.8K]3 years ago
4 0

Answer:

The equation is y = 29x + 90. So, if you plug in 3,100.00 for y.

3,100.00 = 29x + 90

3,100.00 - 90 = 29x

3,010.00 = 29x

3,010.00/29 = 29x/29

x = 103 guests

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Plz help me ASAP! Answer the question in the picture below. Explain all of your steps.
ruslelena [56]

Answer:

175 hope this helps


Step-by-step explanation:

Add 72 and 113

then subtract that from 360


8 0
3 years ago
Only 18 people are allowed on a ride at the same time. There are 157 people waiting in line. How many groups of riders will ther
IrinaK [193]

Answer:

Step-by-step explanation:

Divide 157 by 8 to see how many groups there are.

157/18=8r13. So you can round up to 9 groups of riders as there is a max of 18 at a time.

7 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

<u>e)</u>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
6•10-2 is how many times as large as 2•10-8
pantera1 [17]

Answer:

4 5/6

Step-by-step explanation:

since 6x10-2 is 58, and 2x10-8 is 12, divide 58 by 12 to get 4 5/6

6 0
3 years ago
Read 2 more answers
6. Choose the correct answer.
Gwar [14]

Answer:

d.

Step-by-step explanation:

1) to re-write the given equation from 2y=4x-6 into y=2x-3, then

2) to find the equation like y=2x±... among the a/b/c/d: it is d. y=2x+14. Then

3) to check the given point (-3;8) of the y=2x+14; ⇒ 8=8.

4) correct answer is d. y=2x+14.

6 0
2 years ago
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