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3 years ago

15

A diesel train traveled to the repair yards and back. It took two hours longer to go there than it did to come back. The average

speed on the trip there was 70 km/h. The average speed on the way back was 80 km/h. How many hours did the trip there take?
A) 15
B) 16
C)25
D) 10

1 answer:

erastovalidia [21]3 years ago

7 0

The answer is B 16hrs

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Mr.Daniels left at 2:15 pm and arrived at 11:50 pm what was his travel time

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Twenty-five students from Harry High School were accepted at Magic University. Of those students, 10 were offered athletic schol

Taya2010 [7]

Answer:

Step-by-step explanation:

Part A

For Athletic scholarship,

Mean = (16 + 24 + 20 + 25 + 24 + 23 + 21 + 22 + 20 + 20)/10 = 21.5

Standard deviation = √(summation(x - mean)²/n

n = 10

Summation(x - mean)² = (16 - 21.5)^2 + (24 - 21.5)^2 + (20 - 21.5)^2 + (25 - 21.5)^2 + (24 - 21.5)^2 + (23 - 21.5)^2 + (21 - 21.5)^2 + (22 - 21.5)^2 + (20 - 21.5)^2 + (20 - 21.5)^2 = 64.5

Standard deviation = √64.5/10 = 2.54

For non athletic scholarship,

Mean = (23 + 25 + 26 + 30 + 32 + 26 + 28 + 29 + 26 + 27 + 29 + 27 + 22 + 24 + 25)/15 = 26.6

n = 15

Summation(x - mean)² = (23 - 26.6)^2 + (25 - 26.6)^2 + (26 - 26.6)^2 + (30 - 26.6)^2 + (32 - 26.6)^2 + (26 - 26.6)^2 + (28 - 26.6)^2 + (29 - 26.6)^2 + (26 - 26.6)^2 + (27 - 26.6)^2 + (29 - 26.6)^2 + (27 - 26.6)^2 + (22 - 26.6)^2 + (24 - 26.6)^2 + (25 - 26.6)^2 = 101.6

Standard deviation = √101.6/15 = 2.6

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let 1 be the subscript for scores of athletes and 2 be the subscript for scores of non athletes.

Therefore, the population means would be μ1 and μ2

The random variable is x1 - x2 = difference in the sample mean scores of athletes and non athletes.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

From the information given,

x1 = 21.5

x2 = 26.6

s1 = 2.54

s2 = 2.6

n1 = 10

n2 = 15

t = (21.5 - 26.6)/√(2.54²/10 + 2.6²/15)

t = - 4.65

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [2.54²/10 + 2.6²/15]²/[(1/10 - 1)(2.54²/10)² + (1/15 - 1)(2.6²/15)²] = 1.2008/0.1039

df = 12

We would determine the probability value from the t test calculator. It becomes

p value = 0.00056

Since alpha, 0.1 > than the p value, 0.00056, then we would reject the null hypothesis.

Therefore, these data provide convincing evidence of a difference in ACT scores between athletes and nonathletes.

Part B

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

For a 90% confidence level, the z score from the normal distribution table is 1.645

x1 - x2 = 21.5 - 26.6 = - 5.1

√(s1²/n1 + s2²/n2) = √(2.54²/10 + 2.6²/15) = 1.05

The confidence interval is - 5.1 ± 1.05

This analysis provides evidence that the mean scores for non athletes is higher than the mean scores for athletes, and that the difference between means in the population is likely to be between 4.05 and 6.15

4 0

3 years ago

The equation of function h is h... PLEASE HELP MATH

Flura [38]

Answer:

Part A: the value of h(4) - m(16) is -4

Part B: The y-intercepts are 4 units apart

Part C: m(x) can not exceed h(x) for any value of x

Step-by-step explanation:

Let us use the table to find the function m(x)

There is a constant difference between each two consecutive values of x and also in y, then the table represents a linear function

The form of the linear function is m(x) = a x + b, where

a is the slope of the function
b is the y-intercept

The slope = Δm(x)/Δx

∵ At x = 8, m(x) = 2

∵ At x = 10, m(x) = 3

∴ The slope = $\frac{3-2}{10-8}=\frac{1}{2}$

∴ a = $\frac{1}{2}$

- Substitute it in the form of the function

∴ m(x) = $\frac{1}{2}$ x + b

- To find b substitute x and m(x) in the function by (8 , 2)

∵ 2 = $\frac{1}{2}$ (8) + b

∴ 2 = 4 + b

- Subtract 4 from both sides

∴ -2 = b

∴ m(x) = $\frac{1}{2}$ x - 2

Now let us answer the questions

Part A:

∵ h(x) = $\frac{1}{2}$ (x - 2)²

∴ h(4) = $\frac{1}{2}$ (4 - 2)²

∴ h(4) = $\frac{1}{2}$ (2)²

∴ h(4) = $\frac{1}{2}$ (4)

∴ h(4) = 2

∵ m(x) = $\frac{1}{2}$ x - 2

∴ m(16) = $\frac{1}{2}$ (16) - 2

∴ m(16) = 8 - 2

∴ m(16) = 6

- Find now h(4) - m(16)

∵ h(4) - m(16) = 2 - 6

∴ h(4) - m(16) = -4

Part B:

The y-intercept is the value of h(x) at x = 0

∵ h(x) = $\frac{1}{2}$ (x - 2)²

∵ x = 0

∴ h(0) = $\frac{1}{2}$ (0 - 2)²

∴ h(0) = $\frac{1}{2}$ (-2)² =

∴ h(0) = 2

∴ The y-intercept of h(x) is 2

∵ m(x) = $\frac{1}{2}$ x - 2

∵ x = 0

∴ m(0) = $\frac{1}{2}$ (0) - 2 = 0 - 2

∴ m(0) = -2

∴ The y-intercept of m(x) is -2

- Find the distance between y = 2 and y = -2

∴ The difference between the y-intercepts of the graphs = 2 - (-2)

∴ The difference between the y-intercepts of the graphs = 4

∴ The y-intercepts are 4 units apart

Part C:

The minimum/maximum point of a quadratic function f(x) = a(x - h) + k is point (h , k)

Compare this form with the form of h(x)

∵ h = 2 and k = 0

∴ The minimum point of the graph of h(x) is (2 , 0)

∵ k is the minimum value of f(x)

∴ 0 is the minimum value of h(x)

∴ The domain of h(x) is all real numbers

∴ The range of h(x) is h(x) ≥ 2

∵ m(8) = 2

∵ m(14) = 5

∵ h(8) = $\frac{1}{2}$ (8 - 2)² = 18

∵ h(14) = $\frac{1}{2}$ (14 - 2)² = 72

∴ h(x) is always > m(x)

∴ m(x) can not exceed h(x) for any value of x

<em>Look to the attached graph for more understand</em>

The blue graph represents h(x)

The green graph represents m(x)

The blue graph is above the green graph for all values of x, then there is no value of x make m(x) exceeds h(x)

7 0

3 years ago