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4 years ago

How do you know that 21 over 30 is greater than 2 over 3

Mathematics

2 answers:

fgiga [73]4 years ago

7 0

I know that 21 over 30 is greater than 2 over 3 because 3 is bigger that 2 and the 2 is less than

NISA [10]4 years ago

6 0

21 over 30 is greater than 2 over 3 because 2 over 3 times 10 is 20 over 30. And 21 over 30 is greater than 21 over 30.

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Greg build a new pond which has a volume of 7.35 m it is 4.2 m long and 50 cm deep what is the width of the pond in meters

lozanna [386]

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3 years ago

What goes up but never comes down?

zepelin [54]

Answer:

your age and student debt

Step-by-step explanation:

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2 years ago

Read 2 more answers

How many different 2-digit numbers are there with the following property: the tenth digit is greater than the units digit?

EastWind [94]

Answer:

There is 45 different 2-digit numbers.

Step-by-step explanation:

My way is kinda dumb, but it still works. So, 2 digit numbers is from 10-99. We can start from the 10-19 first.

10, 11,12,13,14,15,16,17,18,19

We need to find the numbers that the tenth digit is greater than the units digit.

10, 11,12,13,14,15,16,17,18,19

Since 1 is the tenth digit, all the ones digits are all going to be bigger. Same goes with 20-29. Then you will have 2 numbers that the tenth digit is greater. My way applies when the tenth digit is only less than the ones digit, not greater. If you do this way...

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

10-19: 1 number

20-29: 2 numbers

30-39: 3 numbers

40-49: 4 numbers

50-59: 5 numbers

60-69: 6 numbers

70-79: 7 numbers

80-89: 8 numbers

90-99: 9 numbers

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So, if you add 1+2+3+4+....+9(which all of you probably memorized by now) it would be 45. The answer is 45. Hope this helped!

8 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

HELP QUICK PLEASE!!!!

Simora [160]

Answer:

41.8°, 138.2° and 401.8°

Step-by-step explanation:

Given the expression;

$3sin^2x + 4sinx - 4 = 0$

Let P = sinx

The expression becomes;

3P²+4P - 4 = 0

Factorize

3P²+6P-2P - 4 = 0

3P(P+2)-2(P+2) = 0

3P-2 = 0 and P+2 = 0

P = 2/3 and -2

When P = 2/3

sinx = 2/3

x = arcsin 2/3

x = arcsin 0.6667

x = 41.8 degrees

Also if P = -2

sinx = -2

x = arcsin (-2)

x will not exist in this case

To get other values of x

sin is positive in the second quadrant

x = 180 - 41.8

x = 138.2°

x = 360+41.8

x = 401.8°

Hence the values of x within the interval are 41.8°, 138.2° and 401.8°

7 0

3 years ago