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Thepotemich [5.8K]
3 years ago
11

Which system of equations is equivalent to the system of equations shown? 2x + 5y = -2 4x + 7y = 8

Mathematics
1 answer:
Elis [28]3 years ago
3 0

Answer:

x=9, y=-4. (9, -4).

Step-by-step explanation:

2x+5y=-2

4x+7y=8

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

-2(2x+5y)=-2(-2)

4x+7y=8

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

-4x-10y=4

4x+7y=8

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

-3y=12

y=12/-3

y=-4

2x+5(-4)=-2

2x-20=-2

2x=-2+20

2x=18

x=18/2

x=9

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3 years ago
Solve 5x - 10 <_ 20 pls help
balandron [24]

Answer:

(-∞, 6]

Step-by-step explanation:

Step 1: Write inequality

5x - 10 ≤ 20

Step 2: Solve for <em>x</em>

  1. <u>Add 10 to both sides:</u> 5x ≤ 30
  2. <u>Divide both sides by 5:</u> x ≤ 6

Here we see that any value less than or equal to 6 would work.

∴ the 3rd option is our answer.

6 0
3 years ago
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34kurt
Your answer is going to be 7.8m.
6 0
2 years ago
Solve the system of equations.
KatRina [158]
Solve by Substitution
Answer:
x = -5
y = -2
5 0
2 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
3 years ago
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