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Hatshy [7]
2 years ago
13

8th GRADE MATH

Mathematics
2 answers:
MrRa [10]2 years ago
4 0

Answer:

It's C

Step-by-step explanation:

sorry if I'm wrong:) but thx

Julli [10]2 years ago
3 0

Answer:

i think its C. sorry if its incorrect

Step-by-step explanation:

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PLZZ HELP!!! NEED ANSWER QUICKLY!!!!
RSB [31]

Answer:

The value of g(x) at x  = -3 is 18.

Step-by-step explanation:

Here, the given function is:

g(x) = - 5 x + 3

Now, for the value of x = -3 ,

g (-3) : Substitute the value of x  = -3 in g(x) , we get

g (-3)  =  -5 (-3) + 3

          =  15 + 3 = 18

or, g ( -3) = 18

Hence, the value of g(x) at x  = -3 is 18.

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4 years ago
The local video store is having a sale on video games. The store is offering a sale pack of 4 video games for $48. What is the p
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A single video game would cost $12
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3 years ago
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Which ordered pair is a solution to the system of inequalities? y <3 and y >-x+5 ?
Aleksandr-060686 [28]
It should be A 6,1 because 6 is the x-axis and 1 is the y-axis
7 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
Which of the following point is located on the line represented by the equation y + 4 = -5 (x - 3)?. .
horsena [70]
If we plug in: x = 3, y= - 4;
- 4 + 4 = - 5 · ( 3 - 3 )
0 = 0 ( correct )
Answer: C ) ( 3, - 4 )

3 0
3 years ago
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