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3 years ago

The ages of two groups of karate students are shown in the following dot plots:

Mathematics

2 answers:

Degger [83]3 years ago

8 0

Answer:

B: Group A has less variability in the data.

Step-by-step explanation:

Group A's ages are around the same area while Group B's age is more spread out.

krok68 [10]3 years ago

6 0

Answer: Group A has less variability in the data.

Step-by-step explanation: Hope this helps <3 :)

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I need help with this because I don’t understand

GalinKa [24]

Answer:

x=1+√5 or x=1−√5

Step-by-step explanation:

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3 years ago

Read 2 more answers

Elena , Matthew, and Kevin painted 5/9 of the wall and Matthew painted 3/9 of the wall and. Kevin painted the rest of the wall.

photoshop1234 [79]

Matthew painted 2/9 of the wall

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3 years ago

6x^2 + 66x + 60 factor completely

kiruha [24]

Hi! Let me help you!

Remember, factoring looks like so:
ab+ac=a(b+c)
So, we need to find the greatest common factor (G.C.F.) of the polynomial.
In this case, 6 is the GCF, so we factor it out:
6x²+66x+60
Divide 6, 66 and 60 by 6:
x²+11x+10

<u>Answer:</u>

x²+11x+10

Hope you find it helpful.

$\rm{Stargazing{\bold{WithJoy}$

3 0

2 years ago

Round 23.546 to the nearest tenth and the nearest hundredth. Explain your thinking.

marysya [2.9K]

Answer:

nearest tenth is 23.5

nearest hundredth is 23.55

Step-by-step explanation:any number more than 5 is rounded up to +1 for the next number so for tenth 23.546= 23.5 and for hundredth 23.546=23.55

5 0

2 years ago

Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.

grandymaker [24]

$(2x+1)^{\cot x}=\exp\left(\ln(2x+1)^{\cot x}\right)=\exp\left(\cot x\ln(2x+1)\right)=\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)$

where $\exp(x)\equiv e^x$ .

By continuity of $e^x$ , you have

$\displaystyle\lim_{x\to0^+}\exp\left(\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)$

As $x\to0^+$ in the numerator, you approach $\ln1=0$ ; in the denominator, you approach $\tan0=0$ . So you have an indeterminate form $\dfrac00$ . Provided the limit indeed exists, L'Hopital's rule can be used.

$\displaystyle\exp\left(\lim_{x\to0^+}\dfrac{\ln(2x+1)}{\tan x}\right)=\exp\left(\lim_{x\to0^+}\dfrac{\frac2{2x+1}}{\sec^2x}\right)$

Now the numerator approaches $\dfrac21=2$ , while the denominator approaches $\sec^20=1$ , suggesting the limit above is 2. This means

$\displaystyle\lim_{x\to0^+}(2x+1)^{\cot x}=\exp(2)=e^2$

7 0

3 years ago