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trasher [3.6K]
3 years ago
11

Is it more difficult to run up a ramp with a slope of 1/5 or a ramp with a slope of 5? Explain.

Mathematics
1 answer:
stich3 [128]3 years ago
6 0

Is it more difficult to run up a ramp with a slope of 1/5 or a ramp with a slope of 5? Explain.  

a dlope of 1/5 is walkable; a slope of 5 is not without climbing gear.

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Write the following numbers in increasing order: −1.4; 2; −3 1 2 ; −1; − 1 2 ; 0.25; −10; 5.2
Pani-rosa [81]

Answer:

-12,-10,-3,-1.4,-1,0.25,2,5.2,12

Step-by-step explanation:

The following number −1.4; 2; −3 1 2 ; −1; − 1 2 ; 0.25; −10; 5.2 in increasing order

-12,-10,-3,-1.4,-1,0.25,2,5.2,12

It's arranged this way starting from the negative sign because positive it's greater than negative and if the negative gets to approach zero it's get smaller

6 0
3 years ago
Find the volume of a rectangular prism with a length of 13.4 ft, a width of 10.3 ft and
iogann1982 [59]

Answer:

690.1

Step-by-step explanation:

13.4 x 10.3 x 5

7 0
2 years ago
Read 2 more answers
In a Geometry class, Rosalie draws two circles on the coordinate grid as shown. She asks her students to prove that the two circ
nataly862011 [7]

Answer:huhhh

Step-by-step explanation:

Yh

5 0
3 years ago
A laboratory technician needs to make a 40-liter batch of a 40% acid solution. how can the laboratory technician combine a batch
Kobotan [32]
You need to do is 40 times 40%
5 0
3 years ago
In tests of a computer component, it is found that the mean time between failures is 937 hours. A modification is made which is
VladimirAG [237]

Answer:

Null hypothesis is \mathbf {H_o: \mu > 937}

Alternative hypothesis is \mathbf {H_a: \mu < 937}

Test Statistics z = 2.65

CONCLUSION:

Since test statistics is greater than  critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

P- value = 0.004025

Step-by-step explanation:

Given that:

Mean \overline x = 960 hours

Sample size n = 36

Mean population \mu = 937

Standard deviation \sigma = 52

Given that the mean  time between failures is 937 hours. The objective is to determine if the mean time between failures is greater than 937 hours

Null hypothesis is \mathbf {H_o: \mu > 937}

Alternative hypothesis is \mathbf {H_a: \mu < 937}

Degree of freedom = n-1

Degree of freedom = 36-1

Degree of freedom = 35

The level of significance ∝ = 0.01

SInce the degree of freedom is 35 and the level of significance ∝ = 0.01;

from t-table t(0.99,35), the critical value = 2.438

The test statistics is :

Z = \dfrac{\overline x - \mu }{\dfrac{\sigma}{\sqrt{n}}}

Z = \dfrac{960-937 }{\dfrac{52}{\sqrt{36}}}

Z = \dfrac{23}{8.66}

Z = 2.65

The decision rule is to reject null hypothesis   if  test statistics is greater than  critical value.

CONCLUSION:

Since test statistics is greater than  critical value; we reject the null hypothesis. Thus, there is sufficient evidence to support the claim that the modified components have a longer mean time between failures.

The P-value can be calculated as follows:

find P(z < - 2.65) from normal distribution tables

= 1 - P (z ≤ 2.65)

= 1 - 0.995975     (using the Excel Function: =NORMDIST(z))

= 0.004025

6 0
3 years ago
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