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musickatia [10]
2 years ago
5

Vugar has a maximum of 30 batteries for toys. Each toy helicopter uses the same number of batteries, and each toy car uses the s

ame number of batteries. Let H represent the number of toy helicopters and C represent the number of toy cars that Vugar can play with without running out of batteries. 3H+4C?303 According to the inequality, how many batteries does each toy helicopter use, and how many batteries does each toy car use? Does Vugar have enough batteries to play with 5 toy helicopters and 4 toy cars? Choose 1 answer: A) Yes B) No
Mathematics
2 answers:
Leno4ka [110]2 years ago
6 0

Answer: 3H+4C<_ 30

The coefficients of the variables H and C represent the number of batteries each toy helicopter and each toy car uses.

Step-by-step explanation:

Each toy helicopter uses 3 batteries, and each toy car uses 4 batteries.

Now let's check whether Vugar has enough batteries for 5 toy helicopters and 4 toy cars. To do this, we substitute  H=5 and C= 4 in the given inequality:

Does Vugar have enough batteries to play with 5 toy helicopters and 4 toy cars?

No, because if you plug in the value for H and C:

3H + 4C<_ 30

3(5) + 4(4) <_30

15 + 16 <_ 30

31 <_ 30; false

Since the inequality is false, Vugar does not have enough batteries to play with 555 toy helicopters and 444 toy cars.

Each toy helicopter uses 333 batteries, and each toy car uses 444 batteries.

No, Vugar does not have enough batteries to play with 555 toy helicopters and 444 toy cars.

Contact [7]2 years ago
5 0

Answer:

Each toy helicopter uses 3 batteries, and each toy car uses 4 batteries.

B) No, Vugar does not have enough batteries to play with 5 toy helicopters and 4 toy cars

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2 years ago
Grades on a standardized test are known to have a mean of 1000 for students in the United States. The test is administered to 45
vovikov84 [41]

Answer:

a. The 95% confidence interval is 1,022.94559 < μ < 1,003.0544

b. There is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. i. The 95% confidence interval for the change in average test score is; -18.955390 < μ₁ - μ₂ < 6.955390

ii. There are no statistical significant evidence that the prep course helped

d. i. The 95% confidence interval for the change in average test scores is  3.47467 < μ₁ - μ₂ < 14.52533

ii. There is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

Step-by-step explanation:

The mean of the standardized test = 1,000

The number of students test to which the test is administered = 453 students

The mean score of the sample of students, \bar{x} = 1013

The standard deviation of the sample, s = 108

a. The 95% confidence interval is given as follows;

CI=\bar{x}\pm z\dfrac{s}{\sqrt{n}}

At 95% confidence level, z = 1.96, therefore, we have;

CI=1013\pm 1.96 \times \dfrac{108}{\sqrt{453}}

Therefore, we have;

1,022.94559 < μ < 1,003.0544

b. From the 95% confidence interval of the mean, there is significant evidence that Florida students perform differently (higher mean) differently than other students in the United States

c. The parameters of the students taking the test are;

The number of students, n = 503

The number of hours preparation the students are given, t = 3 hours

The average test score of the student, \bar{x} = 1019

The number of test scores of the student, s = 95

At 95% confidence level, z = 1.96, therefore, we have;

The confidence interval, C.I., for the difference in mean is given as follows;

C.I. = \left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm z_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore, we have;

C.I. = \left (1013- 1019  \right )\pm 1.96 \times \sqrt{\dfrac{108^{2}}{453}+\dfrac{95^{2}}{503}}

Which gives;

-18.955390 < μ₁ - μ₂ < 6.955390

ii. Given that one of the limit is negative while the other is positive, there are no statistical significant evidence that the prep course helped

d. The given parameters are;

The number of students taking the test = The original 453 students

The average change in the test scores, \bar{x}_{1}- \bar{x}_{2} = 9 points

The standard deviation of the change, Δs = 60 points

Therefore, we have;

C.I. = \bar{x}_{1}- \bar{x}_{2} + 1.96 × Δs/√n

∴ C.I. = 9 ± 1.96 × 60/√(453)

i. The 95% confidence interval, C.I. = 3.47467 < μ₁ - μ₂ < 14.52533

ii. Given that both values, the minimum and the maximum limit are positive, therefore, there is no zero (0) within the confidence interval of the difference in of the means of the results therefore, there is statistically significant evidence that students will perform better on their second attempt after the prep course

iii. An experiment that would quantify the two effects is comparing the result of the confidence interval C.I. of the difference of the means when the student had a prep course and when the students had test taking experience

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Mama L [17]

A bicyclist travels at a constant speed of 12 miles per hour for a total of 45 minutes

Domain x is number of hours

Range y is distance traveled

speed = 12 miles per hour

time = 45 minutes = \frac{45}{60} = 0.75 hours

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Distance = speed * time

Distance = 12 * 0.75= 9 miles

Number of hours x is 0 to 0.75 hours

So domain is { x | 0}

Distance is 0 to 9 miles

So range is { x | 0}

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<span>Using the kinematic equations: (final velocity)^2 = (initial velocity)^2 - 2 * acceleration * distance Assuming the acceleration/deceleration on the car is constant from a constant force on the brakes. Converting from mph to m/s using 0.447 (so 34 mph is 15.2 m/s) (0)^2 = (15.2)^2 - 2 * acceleration * 29 acceleration = 4.0 m/s^2 Had the car been going 105.4 mph (47.1 m/s) (0)^2 = (47.1)^2 - 2 * 4 * distance distance = 277 meters</span>
6 0
2 years ago
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