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julia-pushkina [17]
2 years ago
12

What is the slope of the tine that passes through the points (1/4,3) and (5,8)?

Mathematics
1 answer:
gulaghasi [49]2 years ago
5 0

Answer:

5/4.75 or about 1.05

Step-by-step explanation:

y2-y1 /x2-x1 is the slope formula

8-3 = 5

5-1/4= 4.75

so 5/4.75 or simplified would be about 1.05

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Subduction zones occur at the boundaries between
ycow [4]
D - tectonic plates
Subduction is a geological process that takes place at convergent boundaries of tectonic plates where one plate moves under another and is forced to sink due to high gravitational potential energy into the mantle. Regions where this process occurs are known as subduction zones
7 0
3 years ago
Read 2 more answers
A company is designing a new cylindrical water bottle. The volume of the bottle will be 179 cmcubed. The height of the water bot
tresset_1 [31]

Answer:

The radius of the water bottle is 2.72cm

Step-by-step explanation:

In this question, we are concerned with calculating what the radius of the cylindrical water bottle is given its volume and height.

Mathematically, the volume of a cylinder is V = pi * r^2 * h

From the question, we know that V = 179 cm^3. , h = 7.7cm while r = ?

we substitute these values into the volume equation.

179 = pi * r^2 * 7.7

179 = 3.14 * r^2 * 7.7

179 = 24.178 * r^2

r^2 = 179/24.178

r^2 = 7.403

r = square root 7.403

r = 2.72 cm

7 0
3 years ago
I am not sure how to slove
notsponge [240]

First .... the limit of the given function is NOT EQUAL to 5 (it diverges so it has no limit).  There is a typo.  The 14 should be positive.

\lim_{x \to\ 7} \bigg(\dfrac{x^2-9x+14}{x-7}\bigg)=5

The precise definition of a limit is:

\text{If for every } \epsilon>0\text{ there exists a }\delta >0\text{ such that}\\|f(x)-L|

Given:

f(x) = \dfrac{x^2-9x+14}{x-7}\\\\L=5\\\\a=7\\\\\\|f(x)-L|

⇒ \epsilon = \delta

When ε = 0.1,     δ = 0.1

When ε = 0.01,   δ = 0.01

5 0
3 years ago
The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed
Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

      =\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911

Result is not significant.

3.

(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378

Result is not significant.

4.

(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349

Result is not significant.

5.

(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773

Result is not significant.

6.

(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.

   

8 0
3 years ago
Please help fast........
Neko [114]

Answer:

b

Step-by-step explanation:

8 0
3 years ago
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