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Y_Kistochka [10]
3 years ago
14

Demarco has a piece of fabric 6yd long. He uses a piece 3yd long. He cuts the rest into strips that are each 3\4 yd long. How ma

ny 3\4 yd long strips are there?
Mathematics
1 answer:
Kobotan [32]3 years ago
4 0

Answer: 4

Step-by-step explanation:

Hi, to answer this question, first, we have to subtract the amount of fabric used (3yd) to the total fabric that Demarco has (6yd)

6 -3 =3yd

Finally, we have to divide the result by the length of each strip piece (3/4yd)

3÷ 3/4= 4 pieces  

There are 4 3/4 yd long strips.

Feel free to ask for more if needed or if you did not understand something.

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An animal sanctuary currently has 125 elk in it. The elk are expected to increase by 20% each year. How many elk will there be i
seropon [69]
There will be 200 elk after 3 years
6 0
3 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
Match the equation with the step needed to solve it.
Harman [31]
1 subtract 1 2m - 1 = 3m 
<span>2 subtract 2 2m = 1 + m </span>
<span>3 subtract m m - 1 = 2 </span>
<span>4 add  2 + m = 3 </span>
<span>5 subtract 2m -2 + m = 1 </span>
<span>6 add 1 3 = 1 + m 
1.5.
2.4
3.1
42
56
63</span>
3 0
3 years ago
Solve the compound inequality |3x-9|≤15 and |2x-3|≥5. Give answer in interval notation.
laila [671]

Answer:

The solution of |3x-9|≤15 is [-2;8] and the solution |2x-3|≥5 of is  (-∞,2] ∪ [8,∞)

Step-by-step explanation:

When solving absolute value inequalities, there are two cases to consider.

Case 1: The expression within the absolute value symbols is positive.

Case 2: The expression within the absolute value symbols is negative.

The solution is the intersection of the solutions of these two cases.

In other words, for any real numbers a and b,

  • if |a|> b then a>b or a<-b
  • if |a|< b then a<b or a>-b

So, being |3x-9|≤15

Solving: 3x-9 ≤ 15

3x ≤15 + 9

3x ≤24

x ≤24÷3

x≤8

or 3x-9 ≥ -15

3x ≥-15 +9

3x ≥-6

x ≥ (-6)÷3

x ≥ -2

The solution is made up of all the intervals that make the inequality true. Expressing the solution as an interval: [-2;8]

So, being |2x-3|≥5

Solving: 2x-3 ≥ 5

2x ≥ 5 + 3

2x ≥8

x ≥8÷2

x≥8

or 2x-3 ≤ -5

2x ≤-5 +3

2x ≤-2

x ≤ (-2)÷2

x ≤ -2

Expressing the solution as an interval: (-∞,2] ∪ [8,∞)

6 0
3 years ago
Need help pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee<br><br><br>What is (f⋅g)(x)?
Sergio039 [100]

just need to multiply and axpand ?

( {x}^{2}  + 2)( {x}^{3}  - 4x + 2) =  {x}^{2} ({x}^{3}  - 4x + 2) + 2({x}^{3}  - 4x + 2) \\  =  {x}^{5}  - 4 {x}^{3}  + 2 {x}^{2}  + 2 {x}^{3}  - 8x + 4 \\  =  {x}^{5}  - 2 {x}^{3}  + 2 {x}^{2}  - 8x + 4

5 0
3 years ago
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