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lora16 [44]
3 years ago
11

Solve using the relate-to-a-simpler-problem strategy

Mathematics
1 answer:
Gekata [30.6K]3 years ago
7 0
The correct answer is c i am 99% sure
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Which lines in the graph have a slope greater than 1 but less than 2?
lapo4ka [179]

Answer:

Lines 3 and 4

Step-by-step explanation:

Let's examine the slope of each of the given lines to have the exact value and be able to answer the question. We use the definition of: slope=\frac{rise}{run}

Line 1: slope=\frac{rise}{run}=\frac{6}{2} = 3

Line 2: slope=\frac{rise}{run}=\frac{6}{3} = 2

Line 3: slope=\frac{rise}{run}=\frac{6}{4} = 1.5

Line 4: slope=\frac{rise}{run}=\frac{6}{5} = 1.2

Line 5: slope=\frac{rise}{run}=\frac{6}{6} = 1

Therefore the lines that have slope strictly greater than 1 and less than 2 are:

Lines 3 and 4

3 0
3 years ago
Read 2 more answers
2⁄4 + 3⁄5 + 6⁄8 = ???
denpristay [2]

Answer: 37/20

Step-by-step explanation: hope this helps!

8 0
3 years ago
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Make a box and whisker plot for the data, then identify the shape of the distribution. 1,3,14,9,7,3,6,27,3,13,8,17
vladimir1956 [14]

Answer:

65

Step-by-step explanation:


7 0
3 years ago
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A circle is divided into 9 equal parts. What is the angle measure of two of those parts?
saveliy_v [14]
There are 360 degrees in a circle. That means that we first divide 360 by 9.
360/9=40
We have the measure for one of the nine parts of the circle, but we need two. All we do is multiply 40 by 2 to get 80 degrees.

:)
6 0
3 years ago
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The time college students spend on the internet follows a Normal distribution. At Johnson University, the mean time is 5 hrs wit
netineya [11]

Answer:

The probability that the average time 100 random students on campus will spend more than 5 hours on the internet is 0.5

Step-by-step explanation:

We are given that . At Johnson University, the mean time is 5 hrs with a standard deviation of 1.2 hrs.

Mean = \mu = 5 hours

Standard deviation = \sigma = 1.2 hours

We are supposed to find the probability that the average time 100 random students on campus will spend more than 5 hours on the internet i.e. P(X>5)

Z=\frac{x-\mu}{\sigma}

Z=\frac{5-5}{1.2}

Z=0

P(X>5)=1-P(X<5)=1-P(Z<0)=1-0.5=0.5

Hence the probability that the average time 100 random students on campus will spend more than 5 hours on the internet is 0.5

7 0
3 years ago
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