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amm1812
3 years ago
14

Help please due today and it’s a benchmark

Mathematics
1 answer:
Leona [35]3 years ago
8 0

Answer:

1

Step-by-step explanation:

Ok, so the third and fourth don't seem right. I am going to assume it's either 1 or 2. Sorry if you get it wrong because of me.

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Which number line shows the solution set for in-3155?
IRISSAK [1]

Answer:Find an answer to your question Which number line shows the solution set for in-3155? -8 -6 -4 -2 0 2 4 6 8 -8 -6 -4 -2 0 2 4 6 8 8 6 o -8

Step-by-step explanation:

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
3 years ago
Which is correct classification for the triangle (A) obtuse (B) acute (C)right (D) equiangular
nexus9112 [7]

Answer:

c) Right

Step-by-step explanation:

An Right triangle is a triangle with an angles being 90 degrees and we know that 2 of the angles are 59 and 31 and we know that the sum of all the angles in a triangle will equal 180 so 180-59-31 = 90 so this triangle has a 90 degree angle making it a right triangle

4 0
3 years ago
Express the given terminating decimal as a quotient of integers. if​ possible, reduce to lowest terms. 0.66
taurus [48]
.66 is 66 out of 100.  "out of" means divide

66/100
We can reduce this by dividing each number by the same amount - 2

33/50
8 0
3 years ago
Please help and thank you
d1i1m1o1n [39]
# ov wekz..... \_ >_○ _/
5 0
3 years ago
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