All categories

2 years ago

If AD = 2/3 AB , what is the ratio of the length of arc BC to the length of arc DE

Mathematics

2 answers:

uysha [10]2 years ago

6 0

You have :

--------------

DE arc = ( pi ) ( AD ) ( 2.36 radians / 2 pi radians ) = ( 2/3 ) ( AB ) ( 2.36 radians / 2 )

DE arc = ( 2/3 ( AB ) ( 1.18 radians )

BC arc = ( pi ) ( AB ) ( 1.18 radians / 2 pi radians )

BC arc = ( AB ) ( 0.59 radians )

BC arc / DE arc = ( AB ) ( 0.59 radians ) / ( 2/3 ) ( AB ) ( 1.18 radians )

BC arc / DE arc = ( AB ) ( 0.59 rad ) / ( 2/3 ) ( AB ) ( 1.18 rad )

BC arc / DE arc = ( 3/2 ) ( .59 rad / 1.18 rad ) = 3/4 <-------

ehidna [41]2 years ago

5 0

Answer:

Answer is 3/4

Step-by-step explanation:

Got it right on edmentum

You might be interested in

Archaeologists can determine the diets of ancient civilizations by measuring the ratio of carbon-13 to carbon-12 in bones found

Vsevolod [243]

Answer:

Step-by-step explanation:

The null and the alternative hypothesis are:

H₀=μ₀=μ₁=μ₂=μ₃=μ₄

H₁=two or more μ are different X

Let $X_{ij}$ denote the jth observation of the ith sample. The mean and the variance of the ith sample is given by

$Xbar_{i}$ = $\frac{1}{J_{i}} summation(X_{ij})$ where J=1 to $J_{i}$ and $J_{i}$ is ith sample size

$s_{i} ^{2} =\frac{1}{J_{i}-1 }summation (X_{ij} - Xbar_{i})^{2}$

The sample sizes are J₁ =12 J₂=10 J₃=18 J₄=9

the total number in all samples combined is 49

finding Xbar₁ and s₁

Xbar₁= 1÷12(17.2+....+13.4)

Xbar₁= 17.0500

s₁²= 1÷(12-1) [(17.0500-17.2)+...+(17.0500-13.4)]

s₁²=5.1336

Similarly find the means and variances of other samples

$\left[\begin{array}{ccc}i&Mean&variance\\1&17.0500&5.1336\\2&15.4500&13.2317\\3&16.4972&6.9460\\4&15.4444&7.4428\end{array}\right]$

the sample grand mean denoted by Xbar is the average of all sampled items taken together:

Xbar= $\frac{1}{49}$ (17.2+.....+16.7)

Xbar=16.2255

Compute the treatment sum of squares SSTr, the sum of squares SSE and the total sum of squares SST

SSTr = ∑ $J_{i} (Xbar_{i} -Xbar)^{2}$ from i=1 to 49

SSTr= 20.9910

$SSE= \sum\limits^I_{i=1}\sum\limits^{J_i}_{j=1}(Xbar_{ij}-Xbar_i)^{2}$

$SSE=\sum\limits^I_{i=1}(J_i-1)s_i^{2}$

SSE=(12-1)(5.1336)+...(9-1)(7.4428)

SSE=353.1796

SST=SSTr+SSE

SST=374.1706

Find the treatment mean square MSTr and the error mean square MSE:

MSTr= SSTr/(I-1)

MSTr=6.9970

MSE=SSE/(N-I)

MSE=7.8484

$F=\frac{MSTr}{MSE}$

F=0.89

The degrees of freedom are I-1=3 for the numerator and N-I=45 for the denominator. Under H₀, F has an $F_{3,45}$ distribution. To find the P value we consult the F table.

P>0.100

The complete ANOVA table is below

$\left[\begin{array}{cccccc}source&DF&SS&MS&F&P\\Agegroup&3&20.9910&6.9970&0.89&0.453\\Error&45&353.1796&7.8484\\Total&48&374.1706\end{array}\right]$

(b) The P-value is large. A follower of 5% rule would not reject H₀. Therefore, we cannot conclude the concentration ratios differ among the age groups

4 0

2 years ago

Pls help! i got A is that correct?

In-s [12.5K]

Nope it is incorrect it should be B

3 0

2 years ago

Read 2 more answers

If y = 8x − 2, which of the following sets represents possible inputs and outputs of the function, represented as ordered pairs?

Nataly [62]

You just need to try with all of the pair f.ex
0,-2 -> put 0 in the x-> y=8*0-2=-2 so ok it is q possible pair of the function..
You try every pair till you find set where all the pairs fit
So set 1
1 and 6: 8*1-2=6 ok
2 and 14 : 8*2-2=14 ok
Hope I could help

6 0

2 years ago

Read 2 more answers

Celeste can spend no more than $30 to buy quinoa and rice. She will pay $5 per pound for quinoa and $2 per pound for rice. Which

Furkat [3]

Can you send pic of graph

Step-by-step explanation:

Pls

5 0

2 years ago

Read 2 more answers

Find the area of quadrilateral ABCD. [Hint: the diagonal divides the quadrilateral into two triangles.]

Vadim26 [7]

Answer:

B) 28.53 unit²

Step-by-step explanation:

The diagonal AD divides the quadrilateral in two triangles:

Triangle ABD
Triangle ACD

Area of Quadrilateral will be equal to the sum of Areas of both triangles.

i.e.

Area of ABCD = Area of ABD + Area of ACD

Area of Triangle ABD:

Area of a triangle is given as:

$Area = \frac{1}{2} \times base \times height$

Base = AB = 2.89

Height = AD = 8.6

Using these values, we get:

$Area = \frac{1}{2} \times 2.89 \times 8.6 = 12.43$

Thus, Area of Triangle ABD is 12.43 square units

Area of Triangle ACD:

Base = AC = 4.3

Height = CD = 7.58

Using the values in formula of area, we get:

$Area = \frac{1}{2} \times 4.3 \times 7.58 = 16.30$

Thus, Area of Triangle ACD is 16.30 square units

Area of Quadrilateral ABCD:

The Area of the quadrilateral will be = 12.43 + 16.30 = 28.73 units²

None of the option gives the exact answer, however, option B gives the closest most answer. So I'll go with option B) 28.53 unit²

7 0

2 years ago