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raketka [301]
11 months ago
5

Corey's favorite hiking trail is 10\,\text{km}10km10, start text, k, m, end text long. They wrote this equation to find how much

time it will take to hike the trail (t)(t)left parenthesis, t, right parenthesis if they hike at a rate of rrr kilometers per hour: \dfrac{10}{r}=t r 10 ​ =tstart fraction, 10, divided by, r, end fraction, equals, t Identify the dependent and independent variables.
Mathematics
1 answer:
sammy [17]11 months ago
8 0

The dependent variable and the independent variable are t and r, respectivey

<h3>How to determine the dependent and independent variables</h3>

From the question, we have the following equation that can be used in our computation:

10/r = t

This equation can be re-expressed as

t = 10/r

The above equation implies that t is a function of r

In other words, t is dependent on the variable r

Read more about independent variables at

brainly.com/question/20876341

#SPJ1

<u>Complete question</u>

Corey's favorite hiking trail is 10km. They wrote this equation to find how much time it will take to hike the trail (t) if they hike at a rate of r kilometers per hour: 10/r = t.

Identify the dependent and independent variables.

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The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q
Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

\frac{\sum f}{4}=\frac{100}{4}=25

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF)_{p} = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h

     =34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75

The value of first quartile is 34.75.

Q₃ is at the position:

\frac{3\sum f}{4}=\frac{3\times100}{4}=75

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF)_{p} = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h

     =44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83

The value of third quartile is 44.83.

Then the inter-quartile range is:

IQR = Q_{3}-Q_{1}

        =44.83-34.75\\=10.08

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}

                                 =\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89

Thus, the probability that a student scored at most 50 on the examination is 0.89.

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3 years ago
The differences between the data points and the estimated regression line are called the
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<span>Linear regression is a method of finding the linear equation that comes closest to fitting a collection of data points.
</span>The better the choice of line, the closer the predicted values will be to the observed values.
The differences between the data pints (observed values) and the estimated (pedicted) regression line is called  the <span>residue.
</span>Residue = Observed Value -<span> Predicted Value</span>
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A bowling ball is pushed with a force of 22.0 N and accelerates at 5.5 m/s2. What is the mass of the bowling ball? Ik the answer
KiRa [710]

Step-by-step explanation:

You should always use the least number of significant figures.  The exception is exact numbers or constants.

For example, there are 60 minutes in an hour.  Even though this number has one significant figure, it is an exact number.  So 15 minutes converted to hours would be 0.25 hr.

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3 years ago
Sam is moving bags of soil from a loading dock into a storeroom. The top soil weighs 40 pounds per bag and the mulch weighs 20 p
Levart [38]

Answer:

40x+20y\leq 480

Step-by-step explanation:

Givens

  • The top soil weighs 40 pounds per bag.
  • The mulch weighs 20 pounds per bag.
  • The cart can only carry up to 480 pounds.

Notice that the restriction is a maximum of 480 pounds, that means the inequality must include the sign \leq.

Now, let's call x the top soil and y the mulch, the inequality that represents this problem, would be

40x+20y\leq 480

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3 years ago
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