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julsineya [31]
2 years ago
15

What is the nth term rule of the linear sequence -5,-7,-9,-11,-13

Mathematics
2 answers:
RUDIKE [14]2 years ago
4 0
Your rule for the sequence is -2n-3. Substitute any value for n and you will get the input. Hopefully this helps
monitta2 years ago
3 0

Answer:

-7

Step-by-step explanation:

cause I took the test and got it right in my first try

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Consider the quadratic equation x2=4x-5. How many solutions does the equation have
Norma-Jean [14]
The quadratic formula is =(-b+-sqrt(b^2-4ac))/2a
as you notice the term under the square root is b^2-4ac if it is postive then the equation clearly will have two real soultions if it is negative then the equation will have two imaginary soultion if it is zero then the the equation will have one soultion
so let us calculate b^2-4ac for our given equation
x^2=4x-5 so let us write it in general form which is ax^2+bx+c=0
subtracting 4x from both sides
x^2-4x=-5
adding 5 to both sides
x^2-4x+5=0
a=1,b=-4,c=5
b^2-4ac=(-4)^2-4(1)(5)=16-20=-4
which means the equation has two imaginary soultions
3 0
3 years ago
Read 2 more answers
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
Tell whether (5. 10) is a solution of y = 6x – 18
almond37 [142]

Answer:

NO

Step-by-step explanation:

Simply enter 5, 10 in the x and y boxes to get the answer.

10=6(5)-18

10=30-18

10=12

As a consequence, it isn't a feasible option.

8 0
3 years ago
In the general equation of a line, if A = 0, what will the graph of the line look like?
attashe74 [19]
It would be a zero slope
4 0
3 years ago
Read 2 more answers
HELP NOWW!!!<br> PLSS I NEED HELP RIGHT NOWWW!!<br> PLS HELP
NemiM [27]

Answer:

x = 55

Step-by-step explanation:

Draw a line parallel to the top and bottom parallel lines so this new line goes through the pointed end of x.

Draw another line parallel to the top and bottom lines through the pointy end of 45.

The bottom angle of the line through 45 is 15 degrees (alternate interior angles.

The top angle is 45 - 15 = 30

The bottom angle of the line going to x is 150 degrees. It and the 30 degree angle make 180. 30 + 150 = 180

One final observation The top angle of made by the line going through x is 180 - 25 = 155

What you have now is

155 + 150 + x = 360

305 + x = 360

x = 360 - 305

x = 55

6 0
3 years ago
Read 2 more answers
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