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Liula [17]
3 years ago
10

In six months, 33.6 gallons were used. Find the unit rate.

Mathematics
1 answer:
Nana76 [90]3 years ago
6 0

I believe the answer is 5.6

Just divide 33.6 by 6

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What line segment is equal to AC - BC?<br> (Picture included)
Alexxx [7]

Answer:

<h2><u><em>A_____B</em></u></h2>

Step-by-step explanation:

What line segment is equal to AC - BC?

A_____B_____C -

             B_____C =

A_____B

7 0
2 years ago
write a system of equations for the problem, and then solve the system. if a plane can travel 340 miles per hour with the wind o
BlackZzzverrR [31]

Answer:

w = 40 miles per hour

r = 300 miles per hour

Step-by-step explanation:

We add w with the wind and subtract w against the wind

r+w = 340 miles per hour

r - w = 260 miles per hour

Add the two equations together to eliminate w

r+w = 340 miles per hour

r - w = 260 miles per hour

----------------------------------------

2r = 600

Divide by 2

r = 300 miles per hour without the wind

Now find w

r+w = 340

300 + w = 340

Subtract 300

w = 340-300

w = 40 miles per hour

7 0
4 years ago
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.

5 0
4 years ago
A car travels 220 miles in 4 hours. what is the rate of the car in miles per hour
mixas84 [53]
Answer is 55mph.....
7 0
4 years ago
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Does every number have a decimal expansion?<br><br> yes or no
Aleonysh [2.5K]
I’m pretty sure the answer is no
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3 years ago
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