Rewrite it 2/3 and 5/8 with a denomination of 24

2 Joey runs 4 – miles in of an hour and Chandler runs 2 miles miles in of an hour. Who runs faster

HACTEHA [7]

Answer: Joey is faster

Step-by-step explanation: Because in 1 hour Joey is run 4 mile but in 1 hour Chandler is run 2 mile

6 0

3 years ago

Read 2 more answers

Determine whether the points (–3,–2) and (3,2) are in the solution set of the system of inequalities below. y ≤ ½x + 2 y < –2

lana [24]

Given:

The system of inequalities:

$y\leq \dfrac{1}{2}x+2$

$y$

To find:

Whether the points (–3,–2) and (3,2) are in the solution set of the given system of inequalities.

Solution:

A point is in the solution set of the given system of inequalities if it satisfies both inequalities.

Check for the point (-3,-2).

$-2\leq \dfrac{1}{2}(-3)+2$

$-2\leq -1.5+2$

$-2\leq 0.5$

This statement is true.

$-2$

This statement is also true.

Since the point (-3,-2) satisfies both inequalities, therefore (-3,-2) is in the solution set of the given system of inequalities.

Now, check for the point (3,2).

$2$

This statement is false because $2>-9$ .

Since the point (3,2) does not satisfy the second inequality, therefore (3,2) is not in the solution set of the given system of inequalities.

7 0

3 years ago

Find the indicated term of the given arithmetic sequence. a1 = 115, d = 8, n = 14

olya-2409 [2.1K]

I took the same test and I put the answer D. 1487 and I got it wrong. The correct answer was
A.3

3 0

3 years ago

Which choices are solutions to the following equation?

podryga [215]

You can see solution in the picture, hope that can help you :-)

3 0

4 years ago

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When the author visited Dublin, Ireland (home of Guinness Brewery employee William Gosset, who first developed the t distributio

disa [49]

Answer:

The Decision Rule

Fail to reject the null hypothesis

The conclusion

There is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

Step-by-step explanation:

From the question we are told that

The data is

Car Ages 4 0 8 11 14 3 4 4 3 5 8 3 3 7 4 6 6 1 8 2 15 11 4 1 6 1 8

Taxi Ages 8 8 0 3 8 4 3 3 6 11 7 7 6 9 5 10 8 4 3 4

The level of significance $\alpha = 0.05$

Generally the null hypothesis is $H_o : \mu_1 - \mu_2 = 0$

the alternative hypothesis is $H_a : \mu_1 - \mu_2 > 0$

Generally the sample mean for the age of cars is mathematically represented as

$\= x_1 = \frac{\sum x_i }{n}$

=> $\= x_1 = \frac{4+ 0+ 8 +11 + \cdots + 8
}{27}$

=> $\= x_1 = 5.56$

Generally the standard deviation of age of cars

$\sigma _1 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _1 = \sqrt{\frac{(4 - 5.56)^2 + (0 - 5.56)^2+ (8 - 5.56)^2 + \cdots + 8}{ 27} }$

=> $\sigma _1 = 3.88$

Generally the sample mean for the age of taxi is mathematically represented as

$\= x_2 = \frac{\sum x_i }{n}$

=> $\= x_2 = \frac{8 +8 +0 + \cdots + 4
}{20}$

=> $\= x_2 = 5.85$

Generally the standard deviation of age of taxi

$\sigma _2 = \sqrt{\frac{\sum (x_i - \= x)^2}{n_1} }$

=> $\sigma _2 = \sqrt{\frac{(8 - 5.85)^2 + (8 - 5.85)^2+ (0 - 5.85)^2 + \cdots + 8}{ 20} }$

=> $\sigma _2 = 2.83$

Generally the test statistics is mathematically represented as

$t = \frac{(\= x_ 1 - \= x_2 ) - 0}{\sqrt{\frac{\sigma^2_1}{n_1} + \frac{\sigma^2_2}{n_2} } }$

=> $t = \frac{(5.56 - 5.85 ) - 0}{\sqrt{\frac{(3.88)^2}{27} + \frac{(2.83)}{20} }}$

=> $t = -0.30$

Generally the degree of freedom is mathematically represented as

$df = n_1 + n_2 -2$

$df = 27 + 20 -2$

$df = 45$

From the t distribution table the $P(t > t )$ at the obtained degree of freedom = 45 is

$P(t > -0.30 ) = 0.61722067$

So the p-value is

$p-value = P(t > T) = 0.61722067$

From the obtained values we see that the p-value > $\alpha$ hence we fail to reject the null hypothesis

Hence the there is no sufficient evidence to support the claim that the mean age of the cars is greater than that of taxi

5 0

3 years ago