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2 years ago

What is the approximate area of the shaded sector in the circle ? please explain too

Mathematics

2 answers:

AlladinOne [14]2 years ago

8 0

Short Answer: B
Remark
The general formula for area of a sector is (k) pi r^2 / 360 where k is the central angle of the arc (they are both the same value 110 degrees).

Formula
Area = k * pi * r^2 / 360

Givens
k = 110 degrees.
pi = 3.14
r = 16 inches.

Sub and Solve
Area = 110 * 3.14 * 16^2 / 360
Area = 110 * 3.14 * 256 / 360
Area = 88422.4 / 360
Area = 245.6 in^2 <<<<< Answer

ipn [44]2 years ago

6 0

The area of any circle is (pi · radius²) .

The area of this COMPLETE circle is (pi · radius²) = 256π inches² .

The complete circle covers 360°.

The area of the sector is (110/360) of the whole circle = (110/360)·(256π) in².

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Bob neff, owner of an automotive dealership, pays one of his salesmen, mike, a $1,300 draw per week plus 6% on all commission sa

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Draw per week = $1,300

Draw for four weeks = 1300 x 4 = $5,200

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2 years ago

Which product is positive?

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(Negative two-fifths) (Negative StartFraction 8 over 9 EndFraction) (one-third) (StartFraction 2 over 7 EndFraction)

Step-by-step explanation:

(Negative two-fifths) (Negative StartFraction 8 over 9 EndFraction) (one-third) (StartFraction 2 over 7 EndFraction) = (-2/5)(-8/9)(1/3)(2/7)

Let's take the fraction and eliminate the negative sign because negative multiply negative= positive

(-2/5)(-8/9)(1/3)(2/7)= 32/945

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A scientist mixes x liters of water into a container that has 10 liters of a mixture that is 12% salt and 88% water. The functio

dsp73

Answer:

110 liters of water

Step-by-step explanation:

Let

x ----> liters of water

f(x) ----> the percent of the sat in the new mixture

we have

$f(x)=\frac{1.2}{x+10}$

For $f(x)=1\%=1/100=0.01$

substitute in the equation

$0.01=\frac{1.2}{x+10}$

solve for x

$x+10=\frac{1.2}{0.01}$

$x+10=120\\x=120-10\\x=110\ L$

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2 years ago

Suppose that we want to investigate whether curfews correlate with differences in grades for students in middle school. We selec

Anton [14]

Answer:

Step-by-step explanation:

Hello!

Given the variables

X: Curfew of middle school students (Categorized Yes/No)

Y: Average grade of middle school students. (Categorized: A, B, C, and D)

The objective is to test if there is an association between both variables, you have to conduct a Chi-Square test of independence.

In the null hypothesis, you state that both variables are independent vs. the alternative hypothesis that the variables are dependent.

The hypotheses for this test are:

H₀: Pij= Pi. * P.j i= (1)Yes, (2)No; j= (1)A, (2)B, (3)C, (4)D

H₁: Te variables are not independent.

α: 0.05

$X^2= sum\frac{(O_{ij}-E_{ij})^2}{E_{ij}} ~~X_{(r-1)*(c-1)}$

Where

Oij is the observed frequency for the i-row and j-column

Eij is the expected frequency for the i-row and j-column

r= number of categories in the rows

c= number of categories in the columns

$X^2_{H_0}= 1.4796703$

p-value: 0.687

The p-value is greater than the significance level, so the decision is to not reject the null hypothesis. So using a 5% significance level, you can conclude that having a curfew and the average grades of middle schoolers are two independent variables.

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2 years ago