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3 years ago

10

I have data that tells me how the value of my car has changed from 2012 until now (2021). What type of data display can I use to

show that?
Оа
Oь
Line Graph
Pictograph
Box-and-Whisker Plot
Stem-and-Leaf Plot
od

1 answer:

dolphi86 [110]3 years ago

7 0

Answer:

6. Line graph

Step-by-step explanation:

It shows how information changes over time

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5,250 as scientific notation

earnstyle [38]

$\text{Hey there!}$

$\text{5.250 as a scientific notation}$

$\bf{The\ answer\ is:}$ $\boxed{\bf{5.25 \times10^3}}}}$

$\text{Because, we had to move your decimal approximately 3 placecs to the left}$

$\rightarrow\tex{= 5250. }$

$\boxed{\text{That's how we know the answer is: \bf{\boxed{5.25 \times10^3}}}}\checkmark$

$\text{Good luck on your assignment and enjoy your day!}$

~ $\frak{LoveYourselfFirst:)}$

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3 years ago

Read 2 more answers

A baseball player went up to bat 500 times in a season. He hit the ball 150 times. Find the rate of balls hit to times at bat. E

dem82 [27]

150/500 = 75/250

Here's one example

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3 years ago

PLZ ANSWER THIS!!! I have no clue.

Advocard [28]

Density =mass/ volume
Volume= density/ mass

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3 years ago

Can somebody help me answer these plz?

Snezhnost [94]

number 2 would be 43

number 3 would be 40

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3 years ago

Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate the standard deviation of th

AURORKA [14]

Answer:

95% confidence interval for the population standard deviation of the lifetimes of the batteries produced by the manufacturer.

(8.889, 11.7106)

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given sample size 'n' = 23

Mean of the sample x⁻ = 10.3

Standard deviation of the sample (s) = 2.4

Level of significance = 0.05

Degrees of freedom = n-1 = 23-1 =22

<u><em>Step(ii)</em></u>:-

95% confidence interval for the population standard deviation of the lifetimes of the batteries produced by the manufacturer.

$(x^{-} - t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } , x^{-} +t_{\frac{\alpha }{2} } \frac{S}{\sqrt{n} } )$

$(10.3 - t_{\frac{0.05}{2} } \frac{2.4}{\sqrt{23} } , 10.3 +t_{\frac{0.05}{2} } \frac{2.4}{\sqrt{23} } )$

(10.3 - 2.8188 (0.50043) , 10.3 + 2.8188(0.50043)

(10.3-1.4106 , 10.3+1.4106)

(8.889, 11.7106)

<u><em>final answer</em></u>:-

95% confidence interval for the population standard deviation of the lifetimes of the batteries produced by the manufacturer.

(8.889, 11.7106)

8 0

3 years ago