A sector with an area of 26pie cm^2 has a radius of 6 cm. What is the central angle measure of the sector in radians?

Answer: 7

Step-by-step explanation: whhhhyyyy

Answer:

the answer is attached to the picture

Answer:

5 times 6 is 30 and 30 times 7 is 210

so 210 is your anser hop that helps

Answer:

210

Step-by-step explanation: First way: 5x6=30x7=210

2nd way/distributive property: 5(6x7) 5(6=30x7=210.

Answer:

B. [tex]f(x)=-(x-4)^{2} -3[/tex]

Step-by-step explanation:

[tex]f(x)=x^{2}[/tex]

The transformations are:

Reflected upon the x axis (x is now negative):

[tex]f(x)=-x^{2}[/tex]

Moved 4 units to the right (instead of just the variable x, you have (x-4)):

[tex]f(x)=-(x-4)^{2}[/tex]

And finally moved 3 units down (so now it's minus 3):

[tex]f(x)=-(x-4)^{2} -3[/tex]

Hope this helps! :]

Answer:

a < -1

Step-by-step explanation:

Okay, so we are given this equation:

t(x) = ax^5 + 2

Lets plug in -2 as x and see what happens.. t(x) can be 66..

[tex]a * (-2)^{5} + 2[/tex] = 66

-32a + 2 = 66

-32a = 64

a = -2

A is less than -1,

a < -1

Answer:

64 * b * b * b * b * b * b

Step-by-step explanation:

Apply exponent to number first, then expand the variable's exponent into repeated multiplication.

An exponent is a symbol that is written above and to the right of a mathematical expression to indicate the operation of raising a power.

The value of (2b²)³ without the exponents is 8 x b x b x b x b x b x b.

A symbol that is written above and to the right of a mathematical expression to indicate the operation of raising to a power.

a², b³

a and b are base.

2 and 3 are exponents.

We have,

(2b²)³

This can be written as:

= 2³ x [tex]b^{2\times3}[/tex]

[ 2 x 2 x 2 = 2³ = 8 ], [ a². a³ = [tex]a^{2\times3}[/tex] ] and [ (a²)³ = [tex]a^{2\times3}[/tex] ]

= 8 x [tex]b^{6}[/tex]

Since we have to write without the exponents we have,

= 8 x b x b x b x b x b x b

Thus the value of (2b²)³ without the exponents is 8 x b x b x b x b x b x b.

Learn more about exponents here:

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Answer:

A piece of tape: 15cm =150mm

=> The number of tape 5mm that could be cut: N = 150/5 = 30

Hope this helps!

:)

Answer:

4mm.

Step-by-step explanation:

Area of a Square is: [tex]A=s^2[/tex]

's' -side length

We are given the area of 16 square mm.

[tex]16=s^2\\\sqrt{16} =\sqrt{s^2}\\\boxed {4=s}[/tex]

The length of each side of the square is 4mm.

Answer:

4 mm

Step-by-step explanation:

Hope this helps

Answer:

the radius

Step-by-step explanation:

Answer:

radius

Step-by-step explanation:

Answer:

  [tex]\log_{19}{361}=2[/tex]

Step-by-step explanation:

We know that taking logarithms performs the transformation ...

  [tex]b^e=x\\\\e\cdot\log{b}=\log{x}\quad\text{take logs}\\\\e=\dfrac{\log{x}}{\log{b}}\quad\text{divide by $\log{b}$}\\\\\log_b{x}=e\quad\text{use the change of base formula}[/tex]

Then for b=19, e=2, x=361, we have

  [tex]\boxed{\log_{19}{361}=2}[/tex]

Answer:

b. Use an AD/AS diagram to show how a decrease in the inflation target can keep inflation and short-run output from starting to rise in 2021. Explain your diagram and what it reveals about monetary policy.

Answer:

938583

Step-by-step explanation:

Answer:

The p-value will be "0.0549".

Step-by-step explanation:

The given values are:

Time, t = 15 minutes

Df, σ = 25-1 = 24

Now,

⇒  [tex]H_{0}:\sigma^2\leq 150[/tex]

and,

⇒  [tex]H_{1}:\sigma^2>150[/tex]

As we know,

Chi square = [tex]\frac{(n-1)s^2}{(\sigma^2)}[/tex]

On putting the values in the above formula, we get

⇒                = [tex]\frac{24\times 15^2}{150}[/tex]

⇒                = [tex]\frac{24\times 225}{150}[/tex]

⇒                = [tex]36[/tex]

Therefore, p-value = 0.0549

The p-value determined > 0.05, the null hypothesis also isn't dismissed at point 0.05.

Answer:

2.8

Step-by-step explanation:

using cosine rule

[tex]s^{2}[/tex] = [tex]1.3^{2} + 1.6^{2} - 2(1.3)(1.6)cos 157[/tex]

s = [tex]\sqrt{8.0793}[/tex] = 2.842 = 2.8

Answer:

Only Figure 1 and Figure 3 Are Similar.

Step-by-step explanation:

Figure 1 and Figure 3 are proportional by 1:4 while the Figure 2 is not proportional to any of them.

Answer:

C. Rectangle

Step-by-step explanation:

It will result in to a rectangle

Answer:

15 is your answer

Step-by-step explanation:

2 x 5 =10

3 x 5 =15

Answer:

Question 1) $0.08 per oz

Question 2) $0.09 per oz

Step-by-step explanation:

You have to divide the price by how many ounces there are.

$1.92 / 24 = $0.08

$1.35 / 15 = $0.09

Answer:

0.08.

0.09

Step-by-step explanation:

Answer:

[tex]-2\dfrac{3}{8}[/tex]

Step-by-step explanation:

Given the expression:

[tex]3\dfrac78 - 6\dfrac14[/tex]

We are to simplify and obtain an equivalent expression.

Step 1: Change to Improper Fractions

[tex]3\dfrac78 - 6\dfrac14=\dfrac{31}{8}-\dfrac{25}{4}[/tex]

Step 2: Take the Lowest Common multiple of the denominators

LCM of 8 and 4 is 8.

Therefore:

[tex]\dfrac{31}{8}-\dfrac{25}{4}=\dfrac{31-2(25)}{8}\\\\\dfrac{31-50)}{8}\\\\=-\dfrac{19}{8}\\\\=-2\dfrac{3}{8}[/tex]

Therefore, an equivalent expression to [tex]3\dfrac78 - 6\dfrac14[/tex] is [tex]-2\dfrac{3}{8}[/tex].

Answer

3  7/8+ (-6  1/4)

Step-by-step explanation:

Answer:

x + 57i

Step-by-step explanation:

x - 7² – 8=0

x -49-8

x - 57 = x + 57(-1) ; i = -1( complex number notation)

x + 57i

Answer:

  5 feet

Step-by-step explanation:

Add 0.25 to both sides of the equation:

  4.75 +0.25 = x - 0.25 +0.25

  5.00 = x

The original height of the water in the pool was 5.00 feet.

Answer:

5 feet

hope this helps

Answer:

936 is the correct option

Step-by-step explanation:

Answer:

Hans deposits $6000 into an account.

=> Principal P = 6000

Simple interest rate 5% per year.

=> Rate R = 5% = 5/100 = 0.05

The formula to calculate the amount of interest after 4 years:

A = P x R x years = 6000 x (5/100) x 4 = 1200$

Hope this helps!

:)

Answer:

Step-by-step explanation:

The significant figures of a number that carry meaningful contribution to its measurement resolution

Answer:

Media: 167.88 cm

Mediana:  167.6 cm

Modo: 166.67 cm

Step-by-step explanation:

Hola!

La variable de interés es:

X: estatura de un alumno de noveno año de educación básica.

1)

Primero debes ordenar los datos de menor a mayor y contar cuantos de ellos corresponden dentro de cada intervalo determinado, por ejemplo, el primer intervalo es:

[160;164)

Los intervalos están definidos con el límite inferior cerrado, es decir que incluye el valor de dicho límite, y el límite inferior abierto, es decir, que ese valor no está incluido en el intervalo.

160,160,160,161,162,163,164,165,165,165,165,166,167,167,167,167,168,168,168,169,170, 170, 170,171,173,173,173,175,175,176.

f(1)= 6 (seis valores de estatura corresponden a este intervalo)

La sumatoria de todas las frecuencias absolutas debe dar por resultado el total de observaciones n= 30

Para el segundo intervalo [164;168)

f(2)= 10

2)

hi representa la frecuencia relativa simple y esta se calcula como fi/n

Por ejemplo para el primer intervalo:

h(1)= f(1)/n= 6/30= 0.20

Esta indica la proporción de que las alturas estén entre 160 y 164 cm.

En porcentaje se expresa como hi*100, para el primer intervalo: 0.20*100)= 20%

Para el segundo intervalo h(2)= f(2)/n= 10/30= 0.33 y su porcentaje es 33%

Como indican la proporción de cada categoría de la distribución, la sumatoria de las frecuencias relativas simples de todas las categorías debe ser 1.

3)

Como lo dice su nombre, esta frecuencia es acumulada y se calcula como la sumatoria de las frecuencias absolutas simples, para el primer intervalo, dado que previo a él no hay "nada" es igual a la frecuencia absoluta simple:

F(1)= f(1)

Para el segundo intervalo, es la frecuencia absoluta simple del primer intervalo más la frecuencia relativa simple del segundo intervalo:

F(2)= f(1) + f(2)= 6 + 10= 16

4)

Esta frecuencia también representa la sumatoria de las frecuencias relativas simples.

H(1)= h(1)= 0.20 como previo al primer intervalo no existe distribución definida, la frecuencia relativa acumulada es igual a la frecuencia relativa simple.

Para el segundo intervalo la frecuencia relativa acumulada es:

H(2)= h(1)+h(2)?= 0.20+0.33= 0.57

Adjunta a la respuesta encontrarás la tabla completa.

5)

Como no específica medidas de tendencia central requeridas, voy a calcular la media, mediana y modo utilizando la tabla.

Media

X[barra]= (∑x'fi)/n= ∑x'*hi

Dónde x' representa la marca de clase de cada intervalo. Para calcular la marca de clase de los intervalos debes realizar un promedio entre sus límites y su valor siempre debe encontrarse dentro de los límites del intervalo. Si no es así, has cometido un error de cálculos:

(Limite inferior + Limite superior)/2

1. [160;164)  x₁'= (160+164)/2= 162

2. [164;168)  x₂'= 166

3. [168;172)  x₃'= 170

4. [172;176)  x₄'= 174

Una vez que calculaste las marcas de clase, puedes calcular la media:

X[barra]= ∑x'*hi= (162*0.20)+(166*0.33)+(170*0.27)+(174*0.20)= 167.88 cm

Mediana:

La mediana es el valor de la variable que divide a la muestra en dos (50%-50%).

Para poder calcularla primero debes identificar su posición, en este tipo de presentación, debes identificar el intervalo en el que se encuentra incluida la mediana.

Para muestras pares, la posición de la mediana se calcula como:

PosMe= n/2= 30/2= 15

Esto significa que la mediana corresponde a la 15va observación de la muestra, observando la columna de las frecuencias absolutas (simples o acumuladas) debes identificar cual es el intervalo de la mediana:

Al segundo intervalo se corresponde una frecuencia acumulada de 16, lo que significa que la posición de la mediana está incluida en este intervalo:

[164;168)

Entonces puedes calcular la mediana como:

[tex]Me= Li + c [\frac{PosMe-F_{(i-1)}}{f_i} ][/tex]

Dónde

Li: es el límite inferior del intervalo de mediana.

c: es la amplitud del intervalo

F₍i₋₁₎: frecuencia absoluta acumulada del intervalo anterior al intervalo mediana

fi: frecuencia absoluta del intervalo mediana

[tex]Me= 164 + 4 [\frac{15-6}{10} ]= 167.6[/tex]

Me= 167.6 cm, como puedes notar, el valor de la mediana se encuentra entre los límites del intervalo.

Modo o Moda:

El modo o la moda de una distribución corresponde al valor más observado, es decir, al valor con mayor frecuencia absoluta simple. Al igual que la media, para calcular el modo primero debes identificar el intervalo que lo contiene. En este caso, el intervalo modal será aquel con la mayor frecuencia absoluta simple.

[164;168)

La fórmula para calcular el modo es:

[tex]Md= Li + c[\frac{(f_{max}-f_{ant})}{(f_{max}-f_{ant})+(f_{max}-f_{post})} ][/tex]

Li: es el límite inferior del intervalo modal

c: es la amplitud del intervalo

[tex]f_{max}[/tex]: es la frecuencia absoluta simple del intervalo modal.

[tex]f_{ant}[/tex]: es la frecuencia absoluta simple del intervalo anterior al intervalo modal.

[tex]f_{post}[/tex]: es la frecuencia absoluta simple del intervalo posterior al intervalo modal.

[tex]Md= 164 + 4[\frac{10-6)}{(10-6)+(10-8)} ]= 164+4[\frac{4}{4+2} ]= 166.67[/tex]

Md= 166.67 cm

¡Espero que tengas un buen día!

Answer:

26

Step-by-step explanation:

Answer:

26

Step-by-step explanation:

sinH= hypotenuse/opposite = (2.9/6.6)

H = sin^-1 ( 2.9/6.6)

​H=26.065≈26

Answer:

B, D

Step-by-step explanation:

The product of a sum and a difference results in a difference of squares.

The product must be of the form (a + b)(a - b) to work.

Answer: B, D

Answer:

(w-2.5)(w+2.5) And (-4v-9)(-4v+9)

Step-by-step explanation:

E2020

Answer:

Larry’s graph is incorrect.He should used a closed circle at -5

Step-by-step explanation:

Answer:

A)Larry’s graph is incorrect. He should have used a closed circle at –5.

Step-by-step explanation:

Answer:

[-2; 1 ] is the new column vector

Step-by-step explanation:

See explanation in the attachment.

The value of a-2b as a column vector is [tex]\left[\begin{array}{c}-2&1\end{array}\right][/tex] .

A matrix is an arrangement of numbers, expressions or symbols arranged in rows and columns as a rectangular array. These rows and columns define the size or dimension of a matrix.

For the given situation,

The matrix is

[tex]a=\left[\begin{array}{c}4&5\end{array}\right][/tex] , [tex]b=\left[\begin{array}{c}3&2\end{array}\right][/tex]

The operation is a - 2b. The matrix becomes

⇒ [tex]a-2b=\left[\begin{array}{c}4&5\end{array}\right] -2\left[\begin{array}{c}3&2\end{array}\right][/tex]

⇒ [tex]a-2b=\left[\begin{array}{c}4&5\end{array}\right] -\left[\begin{array}{c}6&4\end{array}\right][/tex]

⇒ [tex]a-2b=\left[\begin{array}{c}-2&1\end{array}\right][/tex]

Hence we can conclude that the value of a-2b as a column vector is [tex]\left[\begin{array}{c}-2&1\end{array}\right][/tex] .

Learn more about matrices here

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Answer:

314 cm²

Step-by-step explanation:

Area of a circle is A= πr²

pi is 3.14 and radius is 10:

A= 3.14(10)²

A= 314

Answer: i dont know

Step-by-step explanation:

Answer:

Step-by-step explanation:

The mean SAT score is [tex]\mu=600[/tex], we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it [tex]\sigma[/tex]) is

[tex]\sigma=48[/tex]

Next they draw a random sample of n=70 students, and they got a mean score (denoted by [tex]\bar x[/tex]) of [tex]\bar x=613[/tex]

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis [tex]H_0:\bar x \geq \mu[/tex]

- The alternative would be then the opposite [tex]H_0:\bar x < \mu[/tex]

The test statistic for this type of test takes the form

[tex]t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}[/tex]

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

[tex]t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\[/tex]

Answer:

The null hypothesis is that the SAT score is not significantly different for the course graduates.

Alternate hypothesis: there is a significant difference between the SAT score achieved by the course graduates as compared to the non-graduates.

Apply the t-test. The Test Statistic value comes out to be t = 1.738 and the p-value = 0.0844

Since the p-value is larger than 0.05, the evidence is weak and we fail to reject eh null hypothesis.

Hope that answers the question, have a great day!