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stira [4]
3 years ago
14

How many tenths are there in four fifths?

Mathematics
2 answers:
hjlf3 years ago
8 0

\dfrac{4}{5}\div \dfrac{1}{10}=\dfrac{4}{5}\cdot10=8

<u>8</u>

sleet_krkn [62]3 years ago
5 0

So to find how many tenths, or 1/10's, there are in four fifths, or 4/5, divide 4/5 by 1/10.

\frac{4}{5} \div \frac{1}{10}

Now, remember that <u>dividing by a number is equivalent to multiplying by its reciprocal</u>. To find the reciprocal, just flip the numerator and denominator around. In this case, 1/10 would become 10/1. Flip 1/10 to its reciprocal, change the sign to multiplication, and multiply numerators with numerators and denominators with denominators:

\frac{4}{5}\times \frac{10}{1}=\frac{40}{5}=8

<u>In short, there are 8 tenths in four fifths.</u>

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Ex 65 : Étagères , merci d’ avance
Alchen [17]

Answer:

Voir ci-dessous.

Step-by-step explanation:

1.

OA/OD = 36/64 = 9/16

OB/OC = 27/48 = 9/16

m<AOB = m<COD

triangle AOB ~ trianlge DOC

m<BAO = m<CDO

AB || CD

2.

AB/CD = OA/OD

AB/80 = 36/64

AB/80 = 9/16

16AB = 9 × 80

AB = 9 × 5

AB = 45

AB = 45 cm

3.

AB = 45

BC = OC + OB = 48 + 27 = 75

AB² + AC² = BC²

45² + AC² = 75²

AC² = 5625 - 2025

AC = √3600

AC = 60 cm

hauteur = 5 × 2 cm + 4 × 60 cm

hauteur = 10 cm + 240 cm

hauteur = 250 cm

6 0
2 years ago
I need help on this hw
s2008m [1.1K]
3rd choice, because you would need to multiply positive 3 to get a positive 9 positive times a positive if positive. negative times a negative is a negative
8 0
3 years ago
Read 2 more answers
For each part, compare distributions (1) and (2) based on their medians and IQRs. You do not need to calculate these statistics;
umka2103 [35]

Answer:

a) The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Step-by-step explanation:

For any distribution,

The median is the variable at the middle when all the variables in the dsitribution are arranged in ascending or descending order.

The median is the (n+1)/2 th variabke.

where n = size of the distribution or number of variables. For these questions, the sample size is 5, hence, the median will always be the 3rd variable when the variables are arranged in ascending or descending order.

The IQR, known as the inter quartile range is a measure of dispersion for the distribution. Although, it isn't as effective as other measures of dispersion such as the standard deviation because the IQR unlike the the standard deviation isn't responsive to changes in the variables, especially ones at the end of the distribution.

The IQR is simply given mathematically as the third quartile minus the first quartile.

IQR = (Third quartile) - (First quartile)

Third quartile is the 3(n+1)/4 th variable. For a sample of n=5, the third quartile is the 4.5th variable, that is the average of the 4th and 5th variable.

First quartile is the (n+1)/4 th variable. For a sample of n=5, the first quartile is the 1.5th variable, that is the average of the 1st and 2nd variable.

Taking the questions one at a time

a) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,6,7,20

Median = 3rd variable = 6

Third quartile = (7+20)/2 = 13.5

First quartile = (3+5)/2 = 4

IQR = 13.5 - 4 = 9.5

The two distributions have the same median (6) but the 2nd distribution has a larger IQR (9.5 > 4).

b) (1) 1,2,3,4,5

Median = 3rd variable = 3

Third quartile = (4+5)/2 = 4.5

First quartile = (1+2)/2 = 1.5

IQR = 4.5 - 1.5 = 3

(2) 6,7,8,9,10

Median = 3rd variable = 8

Third quartile = (9+10)/2 = 9.5

First quartile = (6+7)/2 = 6.5

IQR = 9.5 - 6.5 = 3

The two distributions have a different median with distribution 2 having a bigger median (8 > 3), but they both have the same IQR (3).

c) (1) 3,5,6,7,9

Median = 3rd variable = 6

Third quartile = (7+9)/2 = 8

First quartile = (3+5)/2 = 4

IQR = 8 - 4 = 4

(2) 3,5,7,8,9

Median = 3rd variable = 7

Third quartile = (8+9)/2 = 8.5

First quartile = (3+5)/2 = 4

IQR = 8.5 - 4 = 4.5

The 2nd distribution has a slightly bigger median (7 > 6) and a slightly bigger IQR (4.5 > 4) than the 1st distribution.

d) (1) 0,10,50,60,100

Median = 3rd variable = 50

Third quartile = (60+100)/2 = 80

First quartile = (0+10)/2 = 5

IQR = 80 - 5 = 75

(2) 0,100,500,600,1000

Median = 3rd variable = 500

Third quartile = (600+1000)/2 =800

First quartile = (0+100)/2 = 50

IQR = 800 - 50 = 750

The 2nd distribution consists of variables that are way bigger than those of the 1st distribution, hence, the median and IQR of the 2nd distribution are about ten folds of that of the 1st distribution.

Hope this Helps!!!

5 0
3 years ago
I need this please help me
rewona [7]
Function a has a rate of change of 2 and function b has a rate of change of .75. So function a has a greater rate of change.
4 0
3 years ago
A machine printed 63 booklets every 6 hours. At this rate, which proportion could be used to determine x, the number of booklets
Karolina [17]
63 books every 6 hours=> 63/6 x booklets in 9 hours => x/9 so, 63/6 = x/9 and your answer is A.
7 0
3 years ago
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