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Vesnalui [34]
3 years ago
12

The data given to the right includes data from 30 ​candies, and 5 of them are red. The company that makes the candy claims that

32​% of its candies are red. Use the sample data to construct a 90​% confidence interval estimate of the percentage of red candies. What do you conclude about the claim of 32​%? Construct a 90% confidence interval estimate of the population percentage of candles that are red. _%
Mathematics
1 answer:
lina2011 [118]3 years ago
4 0
100% expansión explanation
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Describe the graph of 2x + y = 11 and 2x + y = 2. Determine the number of solutions.
Alexandra [31]

Answer:

\mathrm{No\:Solution}

Step-by-step explanation:

By using comparing the left hand side of the two given equations we have:

\begin{bmatrix}2x+y=11\\ 2x+y=2\end{bmatrix}\\\\\begin{bmatrix}11=2\end{bmatrix}\\\\11=2\:\mathrm{\:is\:false,\:therefore\:the\:system\:of\:equations\:has\:no\:solution}

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Step-by-step explanation:

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4 years ago
What is the percent of change from 20 to 33?
Andrew [12]

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65%

Step-by-step explanation:

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3 years ago
Does anyone know how to do this? I’m confused
nikklg [1K]

Answer:

cos(θ)

Step-by-step explanation:

Para una función f(x), la derivada es el límite de  

h

f(x+h)−f(x)

​

, ya que h va a 0, si ese límite existe.

dθ

d

​

(sin(θ))=(  

h→0

lim

​

 

h

sin(θ+h)−sin(θ)

​

)

Usa la fórmula de suma para el seno.

h→0

lim

​

 

h

sin(h+θ)−sin(θ)

​

 

Simplifica sin(θ).

h→0

lim

​

 

h

sin(θ)(cos(h)−1)+cos(θ)sin(h)

​

 

Reescribe el límite.

(  

h→0

lim

​

sin(θ))(  

h→0

lim

​

 

h

cos(h)−1

​

)+(  

h→0

lim

​

cos(θ))(  

h→0

lim

​

 

h

sin(h)

​

)

Usa el hecho de que θ es una constante al calcular límites, ya que h va a 0.

sin(θ)(  

h→0

lim

​

 

h

cos(h)−1

​

)+cos(θ)(  

h→0

lim

​

 

h

sin(h)

​

)

El límite lim  

θ→0

​

 

θ

sin(θ)

​

 es 1.

sin(θ)(  

h→0

lim

​

 

h

cos(h)−1

​

)+cos(θ)

Para calcular el límite lim  

h→0

​

 

h

cos(h)−1

​

, primero multiplique el numerador y denominador por cos(h)+1.

(  

h→0

lim

​

 

h

cos(h)−1

​

)=(  

h→0

lim

​

 

h(cos(h)+1)

(cos(h)−1)(cos(h)+1)

​

)

Multiplica cos(h)+1 por cos(h)−1.

h→0

lim

​

 

h(cos(h)+1)

(cos(h))  

2

−1

​

 

Usa la identidad pitagórica.

h→0

lim

​

−  

h(cos(h)+1)

(sin(h))  

2

 

​

 

Reescribe el límite.

(  

h→0

lim

​

−  

h

sin(h)

​

)(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)

El límite lim  

θ→0

​

 

θ

sin(θ)

​

 es 1.

−(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)

Usa el hecho de que  

cos(h)+1

sin(h)

​

 es un valor continuo en 0.

(  

h→0

lim

​

 

cos(h)+1

sin(h)

​

)=0

Sustituye el valor 0 en la expresión sin(θ)(lim  

h→0

​

 

h

cos(h)−1

​

)+cos(θ).

cos(θ)

5 0
3 years ago
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A triangle can exist with side lengths 15cm, 22cm, and 35cm.<br> A<br> False<br> B<br> True
Advocard [28]

Answer:

The answer is False maybe

8 0
3 years ago
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