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Anit [1.1K]
3 years ago
13

I GIVE U BRAINIESt THINg PLZ HelP ASAP pLzBshiwwjbsshh

Mathematics
1 answer:
Lelu [443]3 years ago
5 0

Answer:

She should of used quotient rule since the beginning but if not its step 2

Step-by-step explanation:

You might be interested in
this value of y is a of​ t(y). because is for all values of​ y, it follows that the value of y found in the previous step corres
Radda [10]

For f(t) = 3 sin(t), the minimum is -3, the maximum is 3.

For additional information:

The  period T of the function y=3sin(4t) is;

The period of y = 3 sin (4t) is (2pi) / 4 = pi / 2

The sine function with amplitude A = 0.75 and period T = 10, is

y = 0.75 sin( (2pi / 10) x )

= 0.75 sin( (pi/5) t)

y(4) = 0.75 sin( (pi/5) (4) )

= 0.75 sin ( (4/5) pi ) = .4408

Drawing the sine and cosine function on the same plot shows

that they are identical except for a horizontal shift.

The cosine

function leads the sine function by a shift of (2pi/4) = pi/2.

For the last part, y(4) = 0.75 cos( (2pi/10) (4) ) = - 0.6067

To learn more

visit : brainly.com/question/4842623

#SPJ4

5 0
2 years ago
The historical walking tour of a town covers a total distance of 5 miles. The map shows that the path of the tour travels 3 inch
kari74 [83]
Since the problem is given all the total walking distance, the first thing we are going to do is find the total distance in the map:
Diastance_{map}=3+4+2+1
Distance_{map}=10
Now that we have the distance in the map, we can establish a ratio between the walking distance and the distance in the map:
\frac{5miles}{10inches} = 0.5 \frac{mile}{inch}

We can conclude that each inch the map represents 0.5 miles of walking distance. 

7 0
3 years ago
Read 2 more answers
A. [1 3
Trava [24]

<em>D. [ 6 10 </em>

<em>-8 0]</em>


<em>would be the correct answer.</em>

7 0
3 years ago
Read 2 more answers
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
The circumference of Anyeline's circle is 12 feet. What is the radius (in feet) of her circle? Round your answer to the nearest
Elina [12.6K]

Answer:

1.9

Step-by-step explanation:

you divide 12 by 2 then divided that by pi

6 0
3 years ago
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