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aleksandrvk [35]
3 years ago
10

Extra points!! A candle is 17 in tall after burning for 3 hours. After 5

Mathematics
1 answer:
PIT_PIT [208]3 years ago
4 0

Answer:

1) h=-t+20  

2) 12 inches tall.

Step-by-step explanation:

1) The equation of the line in Slope-Intercept form is:

y=mx+b

Where "m" is the slope and "b" is the y-intercept.

In this case:

y=h (The height of the candle in inches)

x=t (The time in hours)

Then, we can rewrite it:

h=mt+b

Based on the information provided in the exercise, the line passes through these points:

(3,17) and (5,15)

Then, we can find the slope of the line with the formula m=\frac{y_2-y_1}{x_2-x_1}:

 m=\frac{15-17}{5-3}=-1

Now we need to substitute the slope and one of the points into  h=mt+b  and then solve for "b":

17=(-1)(3)+b\\\\b=17+3\\\\b=20

Substituting values, we get that the a linear equation that models the relationship between the heigth of the candle and the time, is:

h=-t+20  

2) We must substitute t=8 into the linear equation  h=-t+20 in order to find the height of the candle after burning 8 hours:

 h=-(8)+20\\\\h=12  

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Exercise 11.21) Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad ba
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Answer:

Step-by-step explanation:

Hello!

The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.

Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.

Be:

X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)

n₁= 12

X[bar]₁= 20.50 hs

S₁²= 199.76hs² (S₁= 14.13hs)

X₂: Battery duration between charges when the primary usage of the phone is internet applications.

n₂= 10

X[bar]₂= 8.50

S₂²= 33.29hs² (S₂= 5.77hs)

Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)

The parameters of interest are σ₁² and σ₂²

a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1

The hypotheses are:

H₀: σ₁²/σ₂² ≤ 1

H₁: σ₁²/σ₂² > 1

There is no level of significance listed so I've chosen α: 0.05

b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:

S²= \frac{1}{n-1}*[∑X²-(∑X)²/n]

For the first sample:

n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36

S₁²= \frac{1}{11}*[7240.36-(246)²/12]= 199.76hs²

S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs

For the second sample:

n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12

S₂²= \frac{1}{9}*[1022.12-(85)²/10]= 33.2911hs²

S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs

c)

For this hypothesis test, the statistic to use is a Snedecors F:

F= \frac{S_1^2}{S_2^2} *\frac{Sigma_1^2}{Sigma_2^2} ~~F_{n_1-1;n_2-1}

This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:

F_{n_1-1;n_2-1; 1-\alpha }= F_{11;9;0.95}= 3.10

The rejection region is then F ≥ 3.10

F_{H_0}= \frac{199.76}{33.29} * 1= 6.0006

p-value: 0.006

Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.

I hope this helps!

8 0
2 years ago
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