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3 years ago

Ray OB divides ∠COA into two angles of what measure? 40° 30° 35° 20°

Mathematics

2 answers:

Andre45 [30]3 years ago

8 0

Answer: 20

Step-by-step explanation:

I suppose the figure that you do not include in your exercise is the one that I attached.

Looking at the figure, it is obvious that the angle COA is divided into two angles of 20 degrees each, so the answer would be 20.

disa [49]3 years ago

3 0

Answer:

The answer is 20, i got it right on odyssey

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Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog

lorasvet [3.4K]

Answer:

a) Standard Error = 0.010

b) 95% Confidence Interval = (0.0924 , 0.1316)

Step-by-step explanation:

a) The formula for Standard Error = √Sample Proportion (1 - sample proportion)/n

Standard Error = √p (1 - p)/n

We are told in the question that:

In a random sample of 400 people, 112 agree and 288 disagree. Estimate the standard error using 1000 samples

p = x/n

n = 1000 because we were told to use it instead of 400

x = number for people that agree = 112

p = 112/1000

p = 0.112

Standard Error = √p (1 - p)/n

= √0.112 (1 - 0.112)/1000

= √0.112 × 0.888/1000

= √0.099456 /1000

= √0.000099456

= 0.0099727629

Approximately to 3 decimal places = 0.010

Therefore, the standard error is 0.010

b) The Question above also asked that we solve for the 95% Confidence Interval

The formula =

p ± z × Standard Error

p = 0.112

z score for 95% confidence interval = 1.96

Standard Error = 0.010

Confidence Interval =

0.112 ± 1.96 × 0.010

= 0.112 ± 0.0196

0.112 - 0.0196

= 0.0924

0.112 + 0.0196

0.1316

Therefore, the 95% confidence interval = (0.0924 , 0.1316)

8 0

3 years ago

Read 2 more answers

The equation line l is given by the equation y= 7/4x+3. line m goes through the point ( –1,4) and is perpendicular to line l whi

mr_godi [17]

We determine line m as follows:

*First, by theorem we have the following:

$m_1=-\frac{1}{m_2}$

Here m1 & m2 are the slopes of two perpendicular lines. For all lines that are perpendicular that is true, so we calculate the slope of line m using the slope of the function given [Which has a slope of 7/4]:

$m_1=-\frac{1}{\frac{7}{4}}\Rightarrow m_1=-\frac{4}{7}$

So, the slope of line m is -4/7. Now, using this slope and the point (-1, 4) we replace in the following expression:

$y-y_1=m_1(x-x_1)$

Here x1, y1 & m1 are the x-component of the point, the y-component of the point, and the slope of the line respectively, so we replace and solve for y:

$y-4=-\frac{4}{7}(x-(-1))\Rightarrow y-4=-\frac{4}{7}x-\frac{4}{7}$ $\Rightarrow y=-\frac{4}{7}x+\frac{24}{7}$

And that last function of y is the line m.

8 0

1 year ago

Calculate the rate of change.

Tpy6a [65]

Answer:

Step-by-step explanation:

Rise is 10 run is 5 Rate of Change is rise over run.

7 0

2 years ago

Charlie invests £1200 at 3.5% per annum compound interest.

lbvjy [14]

Answer:

Step-by-step explanation:

Using the compound interest formula $A = P(1+\frac{r}{n} )^{nt}$

A = amount compounded after n years

P = principal (amount invested)

r = rate (in %)

t = time (in years)

n = time used to compound the money

Given P = £1200., r = 3.5%, t = 3years, n = 1 year(compounded annually)

$A = 1200(1+0.035)^{3}\\ A = 1200(1.035)^{3}\\ A = 1200* 1.108717875\\A = 1,330.46$

Value of Charlie's investment after 3 years is £1,330.46

6 0

3 years ago

The slope of the line tangent to the curve y^2 + (xy+1)^3 = 0 at (2, -1) is ...?

Lina20 [59]

$\frac{d}{dx}(y^2 + (xy+1)^3 = 0)
\\
\\\frac{d}{dx}y^2 + \frac{d}{dx}(xy+1)^3 = \frac{d}{dx}0
\\
\\\frac{dy}{dx}2y + \frac{d}{dx}(xy+1)\ *3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}(xy)+\frac{d}{dx}1\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{d}{dx}xy+x\frac{d}{dx}y+\frac{dx}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(\frac{dx}{dx}y+x\frac{dy}{dx}+\frac{dx}{dx}0\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0$
$\frac{dy}{dx}2y + \left(y+x\frac{dy}{dx}\right)\ * 3(xy+1)^2 = 0
\\
\\\frac{dy}{dx}2y + \left(3y+3x\frac{dy}{dx}\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) (xy+1)^2 = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + \left(3y+3x\ y'\right) ((xy)^2+2xy+1) = 0
\\
\\2y\ y' + 3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y') = 0
\\
\\3x^2y^3 +6xy^2+3y+(3x y' +6x^2y y'+3x^2y y'+2yy' ) = 0
\\
\\y'(3x +6x^2y+3x^2y+2y ) = -3x^2y^3 -6xy^2-3y

$
$y' = \frac{-3x^2y^3 -6xy^2-3y}{(3x +6x^2y+3x^2y+2y )};\ x=2,y=-1
\\
\\y' = \frac{-3(2)^2(-1)^3 -6(2)(-1)^2-3(-1)}{(3(2) +6(2)^2(-1)+3(2)^2(-1)+2(-1) )}
\\
\\y' = \frac{-3(4)(-1) -6(2)(1)+3}{(6 +6(4)(-1)+3(4)(-1)-2 )}
\\
\\y' = \frac{12 -12+3}{(6 -24-12-2 )}
\\
\\y' = \frac{3}{( -32 )}$

7 0

3 years ago