I need help with my last 2 problems

Answer:

answer is the third choice : 104

Step-by-step explanation:

Inscribed Angle Theorem

If an angle is inscribed in a circle, then the measure of the angle equals one half the measure of its intercepted arc.

so the intercepted arc is TWICE the measure of an angle inscribed in a circle

1. A central angle is an angle with endpoints located on a circle's circumference and vertex located at the circle's center. A central angle in a circle determines an arc.

2. An inscribed angle is an angle formed by three points on the circle's circumference.

Angle at the Center Theorem: An inscribed angle is half of the central angle (if they determine the same arc).

In your case angles:

1. ∠QSR is insribed (determines the arc QR);

2. ∠QTR is central (determines the arc QR).

Then by Angle at the Center Theorem, m∠QTR=2m∠QSR=2·52°=104°. Arc QR has the same measure as central angle QSR.

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Answer:

x=27.80

Step-by-step explanation:

we know that cosA=base/hypotaneous

so,

[tex]cos44 = \frac{20}{x} \\ x = \frac{20}{cos(44)} \\ = \frac{20}{0.7193...} \\ = 27.80 [/tex]

Probability that the lock code consists of all even digits is [tex]1344[/tex] if we choose the [tex]4[/tex] digit lock is randomly from numbers [tex]1,2,3,4,5,6,7,8,9[/tex]

How can we find the probability that the lock code consists of all even digits?

Total numbers are [tex]1,2,3,4,5,6,7,8,9[/tex]

And we choose [tex]4[/tex] digit code

So probability of choosing [tex]4[/tex] numbers from [tex]9[/tex] numbers without reapitation is

[tex]9p_{4} =\frac{9!}{(9-4)!} \\=\frac{9*8*7*6*5!}{5!} \\=3024[/tex]

For All numbers are even , probability the last digit of number is even.

So we have [tex]4[/tex]  even number and if we fix one on last digit and can change other [tex]3[/tex] then the probability is

[tex]4*8P_{3} =4*\frac{8!}{(8-3)!} \\=4*\frac{8*7*6*5!}{5!} \\=1344[/tex]

So the total probability of code lock having all even number is [tex]1344[/tex]

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Answer:

[tex]a_{18}=-70[/tex]

Step-by-step explanation:

[tex]a_n=a_1+(n-1)d[/tex]

[tex]a_1=2, d = -4[/tex]

[tex]a_{18}=-2+(18-1)*(-4)=-2-68=-70[/tex]

Answer:

Step-by-step explanation:

true

Answer:

Step-by-step explanation:

sin(90-a)=cos

cos = angle adjacent

theta=BC/CA

Answer:

The area of the square is 121mm

Step-by-step explanation:

To solve this we need two basic formulas:

P = 4s

A = [tex]s^{2}[/tex]

P = perimeter

s = the length of the side

Since the shape is a square all the sides are the same length. To get the value of 's' we need to isolate:

P = 4s

44 = 4s

44 / 4 = 4s / 4

11 = s

Now plug in the answer for 's' to get the area:

A = [tex]s^{2}[/tex]

A = [tex]11^{2}[/tex]

A = 121

Hope this helps.

Step-by-step explanation:

12 = 2 × 2 × 3

18 = 2 × 3 × 3

56 = 2 ×2 × 2 × 7

Now

Common factor = 2

Remaining factor = 2 × 3 × 2 × 3 × 7

LCM = RF × CF

= 504

hence the lCM of 12 , 18 and 56 is 504...

[tex]...[/tex]

The given sample's 95% confidence interval is 11.6 ± 0.5357. I.e., from the lower bound 11.06 to the upper bound 12.14. Its margin error is 0.5357.

To find the confidence interval C.I,

It is given that,

Sample size n = 225

Sample mean μ = 11.6

Standard deviation σ = 4.1

Since 95% confidence interval, z-score is 1.96

So, the required confidence interval is calculated by,

C. I = μ ± z(σ/√n) or mean ± margin error

On substituting,

C. I = 11.6 ± 1.96(4.1/√225)

     = 11.6 ± 1.96 × 0.2733

     = 11.6 ± 0.5357

So, the lower bound = 11.6 - 0.5357 = 11.065 and

the upper bound = 11.6 + 0.5357 = 12.135.

Thus, the confidence interval is from 11.06 to 12.14, and its margin error is 0.0537 for the population mean number of unoccupied seats per flight.

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The time that the first train would get to Oakengats is 08:15.

The time of arrival of the first train can be gotten from the table.

The first entry has a blank space. This would have been the time of arrival for the train.

After the blank space, the next time is when the fits train got to this destination. That is at 08:15

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Answer:

A.  12 grams Zn

B.  280 g of copper.

Step-by-step explanation:

100 grams of brass

20 grams of zinc

100 grams of brass has 20 grams of Zn.

20/100 = 0.200  Zn   20% Zn  and 80% copper.

A.  (60 g brass)*(0.2) = 12 grams Zn

B.  (350 g brass)*(0.80) = 280 g of copper.

Normal income: 25 $

Last week's income: 25 + 20: 45 $

% = Last week's/normal = 45/25 ×100 =180 %

If [tex]$D_{o,\frac{1}{2}} \frac{1}{2} (x, y)[/tex] exists a dilation of △ ABC, the facts regarding the image

△A'B'C' exist AB exists parallel to A'B'.

Any point in the coordinate plane exists directed by a point (x, y), where the x value exists the position of the point with respect to the x-axis, and the y value exists the position of the point with respect to the y-axis.

A dilation exists a transformation [tex]$D_{o,k} \frac{1}{2} (x, y)[/tex], with center O and a scale factor of k that exists not zero, that maps O to itself and any other point P to P'. The center O exists as a fixed point, P' exists as the image of P, and points O, P, and P' exist on the exact line.

In a dilation of [tex]$D_{o,\frac{1}{2}} \frac{1}{2} (x, y)[/tex], the scale factor, 1/2 exists mapping the original figure to the image in such a manner that the distances from O to the vertices of the image exist half the distances from O to the original figure.

Also, the size of the image exists half the size of the original figure.

If [tex]$D_{o,\frac{1}{2}} \frac{1}{2} (x, y)[/tex] exists a dilation of △ ABC, the facts regarding the image △A'B'C' exist AB is parallel to A'B'.

[tex]$D_{o,k} \frac{1}{2} (x, y) = (\frac{1}{2} x, \frac{1}{2}y)[/tex]

The distance from A' to the origin exists half the distance from A to the origin.

Therefore, the correct answer is option A). △ A'B'C' exist AB is parallel to A'B'.

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The value value of x is at the 40th percentile of the distribution is 0.65.

The standard normal distribution is a normal distribution with a mean of zero and standard deviation of 1. The standard normal distribution is centered at zero and the degree to which a given measurement deviates from the mean is given by the standard deviation

Given:- μ[tex]_X[/tex]=0,σ[tex]_X[/tex]=1

We can standardize the random variable: Z=(X−μx)/σx,

Then the equation is P(X≤ x )=Φ(x−μx/σx)=Φ(x−0/1)=0.4

Inverse function

Ф[tex]\\^{-1}[/tex](0.4) = [tex]\frac{x-0}{1}[/tex]

From the table we get,

Ф[tex]^{-1}[/tex] = 0.65

Thus

[tex]\frac{x-40}{10}[/tex] = 0.65

x -0 = 0.65

x = 6.5

Thus the value value of x is at the 40th percentile of the distribution is 0.65

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The change in the area is represented by the expression ΔA = 9 · x + 4, where x is the original width of the rectangle.

By geometry we know that the area of a rectangle is equal to the product of its width (x) and length (y). Based on the statement, the original rectangle is equal to:

A = x · y

A = x · (2 · x - 3)

And the change in the area is:

A + ΔA = (x + Δx) · (y + Δy)

A + ΔA = x · y + Δx · y + Δy · x + Δx · Δy

ΔA = Δx · y + Δy · x + Δx · Δy

ΔA = (2 · x - 3) · Δx + Δy · x + Δx · Δy

ΔA = 2 · (2 · x - 3) + 5 · x + 10

ΔA = 9 · x + 4

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The parts of the circle shown in the image include, chord, diameter, radius, tangent and center point.

The following are the parts of the circle shown;

Thus, the parts of the circle shown in the image include, chord, diameter, radius, tangent and center point.

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In [tex]15[/tex] diffret days [tex]15/4[/tex] inches of rain fell in April month if every day rained [tex]1/4[/tex] of a inch .

How to find the rain fell in [tex]15[/tex] diffret days ?

We know that rain in every day is [tex]1/4[/tex] of a inch

So for [tex]15[/tex] days

[tex]1/4+1/4+1/4+......15times\\=15/4[/tex]

So in [tex]15[/tex] diffret days in April rain fell is [tex]15/4[/tex] inches.

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The values of x and y that are solutions to the simultaneous equations are; x = -11 and y = -4

We are given the simultaneous equations;

4x - 6y = -20   -----(eq 1)

-2x + 4y = 6     -----(eq 2)

Multiply equation 2 by 2 and eq 1 by 1 to get;

-4x + 8y = 12     -----(eq 3)

Add eq 3 to eq 1 to get;

2y = -8

y = -4

Thus;

4x - 6(-4) = -20

4x + 24 = - 20

4x = -20 - 24

4x = -44

x = -44/4

x = -11

Thus, the values of x and y that are solutions to the simultaneous equations are; x = -11 and y = -4

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Answer:

1/a + 1/b + 1/c = 1 , a,b,c is nonzero

Step-by-step explanation:

bc +ac +ab /abc = 1

bc + ac + ab = abc

a = b = c = 1

Answer:

81

Step-by-step explanation:

A perfect square is multiplying a number with the same number in question. 81 is a perfect square because either 9 or -9 will go into it perfectly, like a square's length and height.

[tex]9^{2} = 81[/tex]

Answer: 81

Step-by-step explanation:

81 can be perfectly square rooted into 9. [tex]\sqrt{81}=9[/tex]

Considering the mean of a discrete distribution, you are expected to earn $0.13 by the end of the game.

The expected value of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.

Since 71% of your rolls are successful and 29% of your rolls are unsuccessful, the distribution of your earnings is:

Hence the expected value is:

E(X) = 1 x 0.71 - 2 x 0.29 = 0.71 - 0.58 = 0.13

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They are parallel because they have the same slope of 2.

The segments are given as;

AB => 6x - 3y = 9

CD => 4x - 2y = 8

We start by making y the subject of both equations

AB => 6x - 3y = 9

-3y = -6x + 9

Divide through by -3

y = 2x - 3

CD => 4x - 2y = 8

-2y = -4x + 8

Divide through by -2

y = 2x - 2

A linear equation is represented as:

y = mx + b

Where m represents the slope of the line.

By comparing the three equations, we have:

m1 = 2

m2 = 2

Parallel lines have equal slopes.

This means that the lines AB and CD are parallel lines

Hence, they are parallel because they have the same slope of 2.

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Answer: They are parallel because they have the same slope of 2.

Step-by-step explanation:vvvvvvvvvvvvvvvvvvvvv

[tex]\textbf{ Heya !}[/tex]

Your Answer Is:-

[tex]\sf{9x\div6}[/tex]

because,

               it has a coefficient of 9 (a number that comes before a variable)

`hope that was helpful to u ~

Answer:

$13.5

Step-by-step explanation:

cost of 1 text message= $1.80÷12 =$0.15

cost of 90 messages = $0.15×90 = $13.5

Answer: $13.50 for 90 text messages

Step-by-step explanation:

$0.15×90 = $13.5

$0.15 is the unit rate for 1 text message cost.

Answer: a, c, e

Step-by-step explanation:

a) True. If x=0, f(0)=c.

b) False. Considering the quadratic formula, we know the solutions are not always x=c.

c) True. This is a standard fact.

d) False. If the discriminant is not positive, there are less than two x-intercepts.

e) True. If you plug into the formula [tex]x=-\frac{b}{2a}[/tex], if b=0, then obviously the x-coordinate of the vertex is also 0.

Answer:

Step-by-step explanation:

43

I dont exactly know if this is correct sorry if its wrong

See below for the solution to each question

From the graph, we can see that:

V(x) have a positive value from x > 0 to x < 5.5

When x = 0 and ≥ 5.5, the values of V(x) are either 0 or negative

Hence, the reasonable domain of V(x) is 0 < x < 5.5

From the graph, the maximum value of V(x) is at

V(2) = 182

Hence, the approximate value of x that gives the maximum volume is 2

This is the interval where V(x) increase as x increase

From the graph, V(x) increase from x > 0 to x = 2

Hence, the approximate values of x where the volume increases are 0 < x ≤ 2

The other function is

B(x) = 140

So, the point of intersection represent where the volume is 140

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The expected value of the number of people selected before and including the first time a woman has a turn is 1.6667 and the standard error is 0.7454.

It should be noted that the expected value will be:

1[p(x = 1)] + 2[p(x = 2)] + 3[p(x = 3)]

= 1(0.5) + 2(0.3333) + 3(0.1667)

= 1.6667

Var(X) = E(X²) - E(X)2

= 3.3335 & (1.6667)²

= 0.555611

The standard error will be:

= ✓0.555611

= 0.7454

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The attached graph represents the scatter plot of the table of values

The table of values is given as:

Variable A 1 4 2 3 5 1 3 7 6 6 2 9 7 5 8

Variable B 6 6 1 6 7 4 4 7 5 3 5 25 21 4 4

Rewrite the table as:

x 1 4 2 3 5 1 3 7 6 6 2 9 7 5 8

y 6 6 1 6 7 4 4 7 5 3 5 25 21 4 4

Next, we plot the values on a graph

See attachment for the scatter plot

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