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Juliette [100K]
3 years ago
13

Someone please help me with this and explain it please.

Mathematics
1 answer:
Yuki888 [10]3 years ago
5 0
<h3>Answer: Choice D) \sqrt{34}, 6, 2\pi, \sqrt{42}</h3>

===========================================

Explanation:

Use a calculator to find the decimal form of each of the following below (values are approximate; rounded to 2 decimal places)

  • 2pi = 6.28
  • sqrt(42) = 6.48
  • sqrt(34) = 5.83

We see that the order of the decimal values is: 5.83, 6, 6.28, 6.48

So the order of the terms, from smallest to largest, is \sqrt{34}, 6, 2\pi, \sqrt{42}

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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
A kennel is moving 160 dogs to a new facility. Each dog has its own crate. The facility manager rents 17 trucks. Each truck hold
Galina-37 [17]

Answer:

17.7 = 18 (when rounded)

Step-by-step explanation:

160 divided by 9 = the amount of trucks.

8 0
4 years ago
Draw a model to subtract 2 2/5 -1 8/10
Igoryamba

See picture for reply and solution steps.

6 0
3 years ago
Pls help me asap pls!!
Ne4ueva [31]

Answer:Its A or C

Step-by-step explanation:

3 0
2 years ago
Find an equation for the surface consisting of all points P in the three-dimensional space such that the distance from P to the
erik [133]

Answer:

x^2 +4y +z = 1

Step-by-step explanation:

Surface consisting of all points P to point (0,1,0) been equal to the plane y =1

given point, p (x,y,z ) the distance from P to the plane (y)

| y -1 |

attached is the remaining part of the solution

8 0
3 years ago
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