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Ratling [72]
4 years ago
14

What is the unit rate of (8,14) as a decimal

Mathematics
1 answer:
PilotLPTM [1.2K]4 years ago
6 0
You would need to calculate the unit rate by dividing the y-value by the x-value. 

14/8 = 7/4 

7/4 in decimal form is 1.75
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At 1:00 pm the water level in a pool is 13 inches. At 1:30pm the water level is 18 inches. At 2:30pm the water level is 28 inche
SashulF [63]
From the given question we come to know of certain number of facts and they are:
At 1:00 PM the water level of the pond was = 13 inches
At 1:30 PM the water level of the pond was = 18 inches
At 2:30 PM the water level of the pond was = 28 inches
From the above given facts we can easily find the amount of water changing every half an hour.
Amount of increase in water from 1:00PM to 1:30 PM = (18 - 13) inches
                                                                                       = 5 inches
Amount of increase in water level from 1:30PM to 2:30PM = (28 -18) inches
                                                                                             = 10 inches
From the above two deductions we can come to the conclusion the the constant rate of change in water level is 5 inches for every half an hour.
4 0
3 years ago
Read 2 more answers
If you know how to solve this, Please answer it. Thank You
Mkey [24]

The answer is in the picture. Click on it.

3 0
3 years ago
Please help me with this!!!
guajiro [1.7K]

Answer: 41

Step-by-step explanation:

9*(-2)*(-2) - 4*(-2) - 3 = 9*4 + 8 - 3 = 36 + 5 = 41

3 0
4 years ago
One morning, John drive 5 hours before stopping to eat. After lunch, he. increased his speed by 10 mph. If he completed a 500-mi
Marrrta [24]
So  he drove 5hrs, before lunch, now, he was going at speed, say "r", after he ate, he sped up by 10mph, so, his rate is whatever "r" is, plus 10, or " r + 10 ", after lunch

we know the whole trip took 10hrs, so, is 5hrs before lunch and 5hrs after lunch

we also know the whole trip was 500miles, so if he drove, say "d" miles before lunch, after lunch he drove the slack after lunch, or " 200 - d "

recall, your d = rt, distance = rate * time

\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before lunch}&d&r&5\\
\textit{after lunch}&500-d&r+10&5
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=5r\\
500-d=(r+10)5\\
----------\\
500-\boxed{5r}=(r+10)5
\end{cases}
\\\\\\
\cfrac{500-5r}{5}=r+10\implies 100-r=r+10

and I'm sure you know what that is
6 0
3 years ago
Pamela called nine repair shops for quotes for auto repairs. The prices are $139, $150, $345, $99, $167, $155, $140, $200, and $
ASHA 777 [7]

Answer:

Step-by-step explanation:

The prices of quoted of auto repairs are as listed below;

$139, $150, $345, $99, $167, $155, $140, $200.

For her to check whether there is an outlier in the data, she needs to find the mean of the set of data. An outlier is a value in a set of a data that varies considerably from other data in a dataset. It may be larger or smaller than other set of datas. An outlier can affect the decision of a set of data to be analysed if nor taken care of.

From the data, the possible outliers are $99 and $345

Mean of the data = sum of all the prices/sample size

xbar = \sum Xi / N

Xi are individual datas

\sum Xi = $139+$150+$345+$99+$167+$155+$140+$200+$160

\sumXi = $1555

Sample size = 9

Mean = $1555/9

Mean = $172.78

Hence the value that would best represent the central tendency is $177.78

5 0
3 years ago
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