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vfiekz [6]
3 years ago
15

Camille's skis are 1 1/2 meters long. how many centimeters is this?

Mathematics
2 answers:
slava [35]3 years ago
6 0
1 m = 100 cm, then 1 1/2 x 100= 150cm
Elza [17]3 years ago
5 0
150 centimeters is equal to 1 1/2 meters
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alexdok [17]
Yes, your answer is correct
5 0
3 years ago
Could someone please help me i really don't understand this
Ray Of Light [21]

Answer:

69

Step-by-step explanation:

8 0
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Pls help pls helppp i will giving breanlist
GaryK [48]

Answer:

[-5, 4) ∪ (4, ∞)

Step-by-step explanation:

Given functions:

f(x)=\dfrac{1}{x-3}

g(x)=\sqrt{x+5}

Composite function:

\begin{aligned}(f\:o\:g)(x)&=f[g(x)]\\ & =\dfrac{1}{\sqrt{x+5}-3} \end{aligned}

Domain: input values (x-values)

For (f\:o\:g)(x) to be defined:

x+5\geq 0 \implies x\geq -5

\sqrt{x+5}\neq 3 \implies x\neq 4

Therefore, -5\leq x < 4  and  x > 4

⇒  [-5, 4) ∪ (4, ∞)

3 0
2 years ago
Use the graph below to find the coordinates of J'after J(-5, 1) was reflected over the linex=-3.
k0ka [10]

Answer:

J's would be (-5,-4) if J was reflected over the line x=3.

7 0
3 years ago
I NEED THIS ASAP, PLEASE HELP!!!
WINSTONCH [101]

Answer: 22.5 ; 5 ; 14

Step-by-step explanation:

Given the dataset:

{20,22,23,24,26,26,28,29,30}

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 9

Q1 = 1/4 (9 + 1)th term

Q1 = 1/4(10) = 2.5

We average the 2nd and 3rd term:

(22 + 23) / 2

45 / 2 = 22.5

B) The interquartile range(IQR) of the dataset :

{62,63,64,65,67,68,68,68,69,74}

IQR = Q3 - Q1

The lower quartile (Q1) = 1/4(n + 1)th term

Where n = number of observations, n = 10

Q1 = 1/4 (10 + 1)th term

Q1 = 1/4(11) = 2.75 term

We take the average of the 2nd and 3rd term:

(63 + 64) / 2

45 / 2 = 63.5

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 10

Q3 = 3/4 (10 + 1)th term

Q3 = 3/4(11) = 8.25 term

We take the average of the 8th and 9th term:

(68 + 69) / 2

137 / 2 = 68.5

IQR = Q3 - Q1

IQR = 68.5 - 63.5

IQR = 5

C) give the dataset :

{7,8,8,9,10,12,13,15,16}

The upper quartile (Q3) = 3/4(n + 1)th term

Where n = number of observations, n = 9

Q3 = 3/4 (9 + 1)th term

Q3 = 3/4(10) = 7.5 term

We take the average of the 7th and 8th term:

(13 + 15) / 2

28 / 2 = 14

7 0
3 years ago
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