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3 years ago

Fifty dived by half minus of forty what is the answer?

Mathematics

2 answers:

strojnjashka [21]3 years ago

8 0

Step-by-step explanation:

50÷1/2-40

50×2-40

100-40

the answer is 60

MariettaO [177]3 years ago

6 0

Answer:60

Step-by-step explanation:if the equation is 50/.5-40, the first step is to turn them into fractions 50/1 and 50/100 (or 1/2) then you flip the second fraction and change the sign to multiply so it would be 50/1x2/1.

Y=50/.5-40

Y=50/1x2/1-40

Y=100-40

That leaves you with 100/1 (or 100). Then you subtract 40 from 100

100-40=60

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Which number line shows the solution of 5x - 25 < -15?

Anit [1.1K]

I think a. But I don’t understand this question.

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3 years ago

The ages of students enrolled in two math classes at the local community college, Class A and Class B, are listed in order below

nlexa [21]

Answer:

The true statement about Class B is that Class B has a smaller median and the same inter quartile range.

Step-by-step explanation:

We are given the ages of students enrolled in two math classes at the local community college, Class A and Class B, below;

Class A: 20, 20, 20, 21, 22, 23, 23, 25, 27, 29, 30, 31, 34, 35, 36, 39, 40

Class B: 16, 17, 18, 18, 20, 22, 22, 24, 26, 26, 28, 29, 30, 34, 37, 40, 42

1) Firstly, we will calculate Median for Class A;

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 27

Hence, the median of class A is 27.

2) Now, we will calculate Median for Class B;

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 26

Hence, the median of class B is 26.

3) Now, we will calculate the Inter quartile range for Class A;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $21+ 0.5[22- 21]$

= 21.5

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $34+ 0.5[35- 34]$

= 34.5

Therefore, Inter quartile range for Class A = 34.5 - 21.5 = 13.

4) Now, we will calculate the Inter quartile range for Class B;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $18+ 0.5[20- 18]$

= 19

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $30+ 0.5[34- 30]$

= 32

Therefore, Inter quartile range for Class B = 32 - 19 = 13.

Hence, the true statement about Class B is that Class B has a smaller median and the same inter quartile range.

4 0

4 years ago

1. The values, x in a sample of 15 are summarized as follows

Serhud [2]

Answer:

(a) 100

(b) 10.27

Step-by-step explanation:

We are given

No of elements = 15

Σ(x-c) = 72,Σ(x-c)^2 = 499.6

,where c is a constant

and the sample mean is 104.8.

(a) lets take into account Σ(x-c) = 72

this means that we have the sum of the 15 elements of x and each element of x is subtracted by the constant c

so the equation becomes Σxi -15c = 72, ............(1)

where xi means the sum of the elements of x from 1 to 15.

we are given the mean as 104.8

this means that Σxi/15 = 104.8

Σxi = 15*104.8 = 1572 .............(2)

substituting (2) in (1)

we get

1572 - 15c = 72

15c = 1500

c = 100

(b) We will use the property that variance does not change when a constant value is added or subtracted to the elements. This we can observe in the given equation that c is a constant that has the value of 100.

so the variance is

σ^2 = Σ(x-c)^2/15 - (Σ(x-c)/15 )^2

= 499.6/15 - (72/15)^2

= 33.31 - 23.04

σ^2 = 10.27

Therefore the variance of the given problem is 10.27.

5 0

3 years ago

The rug is 21 feet long and 20 feet wide. What is the length of the diagonal?

ser-zykov [4K]

Answer:

using hyp²=opp²+adj²