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Alex73 [517]
3 years ago
6

map of a city uses the scale 1 cm equals 50 m. On the​ map, South Street is 25 cm long. If there is a traffic cone at the start

of South Street and a cone every 10 m along the​ road, how many traffic cones are there in​ all?
Mathematics
1 answer:
aivan3 [116]3 years ago
4 0

Answer:

125 traffic cones

Step-by-step explanation:

In order to find the total number of meters, we can set up a proportion to find based on the ration of cm to m:

\frac{centimeters}{meters}=\frac{1}{50}=\frac{25}{x}, where 'x' is the number of meters

cross-multiply: x = 25(50) or x = 1250 m

Since there is a cone every 10 m, we can divide the total number of meters by 10:

1250/10 = 125 cones

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Answer:

I would say A. 7.8 x 10^-5

Step-by-step explanation:

7.8 x 10^-5 can also be written as 7.8 times 10 to the negative fifth power or the exponent of negative five. But, if you solve it you would get 0.000078 as the real number.

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One serving of cottage cheese has a mass 110 grams. 2/55 of the mass consists of carbohydrates. What is the total mass of carboh
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The answer is 8 because 110 divided by 55 multiplied by 2 is one serving and times that by 2 and you get 8

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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!

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