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gizmo_the_mogwai [7]
3 years ago
10

I had four numbers: 12, 8, 4, and ????? However, I remember that their average was 11. Please, help me to remember the number th

at I forgot!!! The number that our old math teacher forgot is
Mathematics
2 answers:
Anestetic [448]3 years ago
5 0

Answer:

The number was 20

Step-by-step explanation:

As the average is 11 and the numbers are 4, so the total sum of numbers will be 11X4=44. now the sum of three numbers are (20+8+4)=24. So the last number will be= (44-24)=20. :)

dolphi86 [110]3 years ago
5 0

Answer:

20, 12,8,4 average out to 11

Step-by-step explanation:

omni average calculator on google will help :)

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Helppppp me please…….:/
Dmitry_Shevchenko [17]

Answer:

65.3%

Step-by-step explanation:

what we want: p(2)+p(3)+p(4)

P(2)={5\choose2}*.4^2*.6^3\\P(3)={5\choose3}*.4^3*.6^2\\P(4)={5\choose4}*.4^4*.6\\P(2)+P(3)+P(4)= .6528

which equal 65.28%

which rounds to

65.3%

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
4 years ago
Does (3,6) make the inequality 4x+y >19 true
mrs_skeptik [129]
(3,6)
4x + y > 19

4(3) + 6 > 19
12 + 6 > 19
18 > 19

No
6 0
4 years ago
There are 20 pieces of fruit in a bowl and 7 of them
makkiz [27]

Answer: 35%

Step-by-step explanation:

3 0
3 years ago
you are allowed to multiply as many 2's and/or as many 5's as you want. What can be the last didget of your result
Levart [38]

Based on the fact that you are allowed to multiply as many 2's and/ or as many 5's as you want, the last digit of your result can be either a 2, 5 or 0.

<h3>What is the last digit of your result?</h3>

If you multiplied a 5 number with another 5 number, you will either get a 0 as the last digit or a 5 as the last digit:

5 x 5 = 25

5 x 10 = 50

5 x 15 = 75

5 x 25 = 125

If you multiply a 5 with a 2, you result would always end in 0:

5 x 2 = 10

5 x 4 = 20

5 x 6 = 30

If you multiply a 2 with another 2, the result would be a 2:

2 x 2 = 4

2 x 4 = 8

2 x 6 = 12

Find out more on operations involving 5s at brainly.com/question/11977981.

#SPJ1

5 0
2 years ago
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