Find the marked price if: Selling price =Rs 4999.50 and Discount percentage = 10%.​

Answer:

The width of the confidence interval based on the 784 women would be narrower than the  confidence interval based on the 327 women.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

From this, we have that the margin of error, and also the width, is inversely proportional to the sample size, that is, a larger sample size leads to a smaller margin of error and a narrower interval.

How would the width of the new confidence interval compare to the width of the  confidence interval based on the 327 women in the armed forces?

New interval: 784

Old interval: 327

Sample size increased, so the new interval will be narrower, and the correct answer is:

The width of the confidence interval based on the 784 women would be narrower than the  confidence interval based on the 327 women.

if you like my answer then please mark me as brainliest

2x+54=8x

So. The Answer is 7

Answer:

2

y = 3x-15

y=5x-35

x1=5

x2=7

distance between = 7-5 = 2

Step-by-step explanation:

Answer:

Mike had 72 stamps

Ken had 120 stamps

Step-by-step explanation:

Given

[tex]M \to Mike[/tex]

[tex]K \to Ken[/tex]

[tex]\frac{1}{5} * K = \frac{1}{3} * M[/tex]

[tex]K + 24 = 3 * ( M - 24)[/tex]

Required

Find K and M

Make K the subject in: [tex]K + 24 = 3 * ( M - 24)[/tex]

[tex]K = 3 * ( M - 24) - 24[/tex]

Substitute [tex]K = 3 * ( M - 24) - 24[/tex] in [tex]\frac{1}{5} * K = \frac{1}{3} * M[/tex]

[tex]\frac{1}{5} * [3 * ( M - 24) - 24] = \frac{1}{3} * M[/tex]

Open brackets

[tex]\frac{1}{5} * [3M - 72 - 24] = \frac{1}{3} * M[/tex]

[tex]\frac{1}{5} * [3M -96] = \frac{1}{3} * M[/tex]

Multiply both sides by 15

[tex]3* [3M -96] = 5 * M[/tex]

[tex]9M -288 = 5M[/tex]

Collect like terms

[tex]9M -5M= 288[/tex]

[tex]4M= 288[/tex]

Divide both sides by 4

[tex]M= 72[/tex]

Substitute [tex]M= 72[/tex] in [tex]K = 3 * ( M - 24) - 24[/tex]

[tex]K = 3 * (72 - 24) - 24[/tex]

[tex]K = 3 * 48 - 24[/tex]

[tex]K = 120[/tex]

Answer:

Linear systems are systems of equations in which the variables are never multiplied with each other but only with constants and then summed up

Answer:

C.

0.5; 13.5; 1093.5

Step-by-step explanation:

Answer:

g/c =m

Step-by-step explanation:

g=4cm-3cm

Combine like terms

g = 1cm

g =cm

Divide by c

g/c = cm/c

g/c =m

area of the parallelogram is: b x h

12 x 4 = 48 (d)

Answer:

67°

Step-by-step explanation:

let the base angle=x

x+x+46=180

2x=180-46=134

x=134/2=67

[tex] \frac{ \sin(20 \degree) }{ \cos(70\degree) } + \frac{ \cos(35\degree) }{ \sin( 65\degree)} [/tex]

[tex] \frac{ \sin(20 \degree) }{ \cos(70\degree) } + \frac{ \cos(35\degree) }{ \sin( 65\degree)} [/tex]

[tex] \sin( \theta) = \cos(90 - \theta) \\ \\ \cos( \theta) = \sin(90 - \theta) [/tex]

So putting down the value

[tex] \frac{ \cos(90 - 20 \degree) }{ \cos(70\degree) } + \frac{ \sin(90 - 35\degree) }{ \sin( 65\degree)} [/tex]

[tex] \frac{ \cos(70\degree) }{ \cos(70\degree) } + \frac{ \sin(65\degree) }{ \sin( 65\degree)} [/tex]

[tex]\frac{\cancel{\cos(70\degree)}}{ \cancel{\cos(70\degree)}} + \frac{\cancel{\sin(65\degree)}}{\cancel{\sin( 65\degree)}} [/tex]

[tex] \frac{1}{1} + \frac{1}{1} \\ 1 + 1 = 2 \: \: ans[/tex]

Answer:

y''=(4-3t)/[16 t^3 (2-t)^(3/2)]

Step-by-step explanation:

x = t², y = (2 - t)^1/2

dy/dx=dy/dt×dt/dx by chain rule

dy/dt=1/2 (2-t)^(1/2-1) × (-1)

dy/dt=-1/2 (2-t)^(-1/2)

dy/dt=-1/[2(2-t)^(1/2) ]

dx/dt=2t

dy/dx=-1/[2(2-t)^(1/2) ] × 1/[2t]

dy/dx=-1/[4t (2-t)^(1/2) ]

We need to find the second derivative now.

That is we calculate d/dt(dy/dx in terms of t) then divide by derivative of x in terms of t).

dy/dx=-1/[4t (2-t)^(1/2) ]

Let's find derivative of this with respect to t.

d/dt(dy/dx)=

[0[4t (2-t)^(1/2)]-(-1)(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [4t (2-t)^(1/2) ]^2

Let's simplify

d/dt(dy/dx)=

[(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [4t (2-t)^(1/2) ]^2

Continuing to simplify

Apply the power in the denominator

d/dt(dy/dx)=

[(4(2-t)^(1/2)+-4t(1/2)(2-t)^(-1/2))]/ [16t^2 (2-t) ]

Multiply by (2-t)^(1/2)/(2-t)^(1/2):

d/dt(dy/dx)=

[(4(2-t)+-4t(1/2)]/ [16t^2 (2-t)^(3/2)]

Distribute/multiply:

d/dt(dy/dx)=

[(8-4t+-2t)]/ [16t^2 (2-t)^(3/2)]

Combine like terms:

d/dt(dy/dx)=

[(8-6t)]/ [16t^2 (2-t)^(3/2)]

Reducing fraction by dividing top and bottom by 2:

d/dt(dy/dx)=

[(4-3t)]/ [8t^2 (2-t)^(3/2)]

Now finally the d^2 y/dx^2 in terms of t is

d/dt(dy/dx) ÷ dx/dt=

[(4-3t)]/ [8t^2 (2-t)^(3/2)] ÷ 2t

d/dt(dy/dx) ÷ dx/dt=

[(4-3t)]/ [8t^2 (2-t)^(3/2)] × 1/( 2t)

d/dt(dy/dx) ÷ dx/dt=

[(4-3t)]/ [16t^3 (2-t)^(3/2)]

Or!!!!!!

x = t², y = (2 - t)^1/2

Since t>0, then t=sqrt(x) or x^(1/2).

Make this substitution into the equation explicitly solved for y:

y = (2 - x^1/2)^1/2

Differentiate:

y' =(1/2) (2 - x^1/2)^(-1/2) × -1/2x^(-1/2)

y'=-1/4(2 - x^1/2)^(-1/2)x^(-1/2)

y'=-1/4(2x-x^3/2)^(-1/2)

Differentiate:

y''=1/8(2x-x^3/2)^(-3/2)×(2-3/2x^1/2)

y''=(2-3/2x^1/2)/[8 (2x-x^3/2)^(3/2)]

Replace x with t^2

y''=(2-3/2t)/[8 (2t^2-t^3)^(3/2)]

Multiply top and bottom by 2

y''=(4-3t)/[16 (2t^2-t^3)^(3/2)]

Factor out t^2 inside the 3/2 power factor:

y''=(4-3t)/[16 (t^2)^(3/2) (2-t)^(3/2)]

y''=(4-3t)/[16 t^3 (2-t)^(3/2)]

Answer:

Is this a trick question ???

Why would the ratio of salt to pepper change if you doubled the recipe?

The amount of salt and pepper would double, but not the ratio...

the answer should be 2:3

Step-by-step explanation:

Answer:

45 degrees

Step-by-step explanation:

the 2 angles are supplementary (add up to 180 degrees)

180-135=45

Answer:

2020³-1/4041

Step-by-step explanation:

9514 1404 393

Answer:

  about 79 ft

Step-by-step explanation:

The relevant trig relation is ...

  Tan = Opposite/Adjacent

In this scenario, the angle is 23°, the adjacent side is the distance to the building, and the opposite side is the building height. Then we have ...

  height = tan(23°)·(186 ft) ≈ 78.95 ft ≈ 79 ft

Answer:

The probability that you won't be a hobbit=

[tex] \frac{2}{3} [/tex]

The probability that you won't choose an umbrella=

[tex] \frac{4}{5} [/tex]

Step-by-step explanation:

The characters are 3, minus the hobbit gives you 2. So the probability of that=

[tex] \frac{2}{3} [/tex]

The weapons are 5, minus the umbrella is 4. So the probability of that=

[tex] \frac{4}{5} [/tex]

The Pascal triangle terms and binomial are related by the binomial theorem fomrula:

The row of the Pascal triangle is the same as

where n is the same as the row of the Pascal triangle.

For example,

.

is the same as the 2nd row of the Pascal triangle.

.

Also know the leading coefficient and last term degree will be the nth row of the Pascal triangle.

The first term, a will be 1 less than the previous terms and the second term, b will be 1 more than the previous terms. For example, look at

.

Another example is

.

The third row of Pascal Triangle

.

.

Pascal's triangle provides a systematic way to determine the coefficients of binomial expansions, making it a valuable tool in algebra and combinatorics.

Binomial expansions are directly related to Pascal's triangle.

Pascal's triangle is a triangular arrangement of numbers in which each number is the sum of the two numbers directly above it. The first and last numbers in each row of Pascal's triangle are always 1.

The coefficients of the terms in a binomial expansion correspond to the numbers in Pascal's triangle. Each row of Pascal's triangle represents the coefficients of the corresponding power of the binomial.

For example, in the expansion of (a + b)², the coefficients are 1, 2, and 1. These coefficients can be found in the third row of Pascal's triangle: 1, 2, 1. Similarly, in the expansion of (a + b)³, the coefficients are 1, 3, 3, and 1, which can be found in the fourth row of Pascal's triangle: 1, 3, 3, 1.

In summary, Pascal's triangle provides a systematic way to determine the coefficients of binomial expansions, making it a valuable tool in algebra and combinatorics.

Learn more about Pascal's triangle here;

https://brainly.com/question/29549939

#SPJ2

Answer:

15

Step-by-step explanation:

AB+BC=AC

2X+3+(4X-11)

6X-8=28

6x= 36

x=6

then ab= 2(6)+3

=15

bc= 4(6)-11

=13

ac=ab+bc

=15+13

=28

Solution :

Case I :

If Collen is late on [tex]0[/tex] out of [tex]5[/tex] days.

[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} $[/tex]

[tex]$=\frac{1}{32}[/tex]

Case II :

When Collen is late on [tex]1[/tex] out of [tex]5[/tex] days.

[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times ^5C_1$[/tex]

[tex]$=\frac{1}{32} \times 5$[/tex]

[tex]$=\frac{5}{32}[/tex]

Case III :

When Collen was late on [tex]2[/tex] out of [tex]5[/tex] days.

[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times ^5C_2$[/tex]

[tex]$=\frac{1}{32} \times 10$[/tex]

[tex]$=\frac{5}{16}[/tex]

Therefore, the [tex]\text{probability}[/tex] that Collen will arrive late to work no more than [tex]\text{twice}[/tex] during a [tex]\text{five day workweek}[/tex] is :

[tex]$=\frac{1}{32} + \frac{5}{32} + \frac{5}{16} $[/tex]

[tex]$=\frac{1}{2}$[/tex]

You can model the ellipse with the equation

(x/6)² + (y/5)² = 1

If you're comfortable with calculus, read on.

Let x = 6 cos(t ) and y = 5 sin(t ), with 0 ≤ t ≤ π/2. Then the curve C parameterized by (x(t), y(t)) traces out one quarter of the ellipse. The arc length of this curve is given by the integral,

[tex]\displaystyle\int_C\mathrm ds = \int_0^{\pi/2}\sqrt{x'(t)^2+y'(t)^2}\,\mathrm dt \\\\ = \int_0^{\pi/2}\sqrt{36\sin^2(t)+25\cos^2(t)}\,\mathrm dt \\\\ = \int_0^{\pi/2}\sqrt{25+11\sin^2(t)}\,\mathrm dt \\\\ = 5\int_0^{\pi/2}\sqrt{1+\frac{11}{25}\sin^2(t)}\,\mathrm dt[/tex]

Unfortunately, you cannot find the exact value of this integral. (Well, you can, but it involves introducing a special kind of function; I'll link it in a comment.)

But you can approximate it, and the integral above comes out to about 8.657.

I'm not sure if there's a simpler, purely geometric way to estimate the arc length. But since the semiaxes (5 ft and 6 ft) are fairly close, you could try a circle with a radius somewhere in between. The arc length would range between 7.854 ft and 9.425 ft. If the material isn't too expensive, you can always overshoot and trim off the excess or repurpose it.

Answer:

The conic section for the equation 5x^2 - y = 12 is parabola

Answer:

the answer is option 1

Step-by-step explanation:

the negative angle => a quarter round angle (clockwise) = ¼ x ( -360°) = -90°

and

the positive angle => a full round angle + 270° = (360°)+270°

= 630°

Answer:

correct answer is -90 and 630

Step-by-step explanation:

Answer:

4x-7=-5x²

or, 5x²+4x-7=0

so the standard form of the equation is,

5x²+4x-7=0

or, f(x) = 5x²+4x-7

Answered by GAUTHMATH

Answer:

[tex]\Large \boxed{x_1=\frac{9+\sqrt{51} }{3} \ \ ; \ \ x_2=\frac{9-\sqrt{51} }{3} }[/tex]

Step-by-step explanation:

[tex]\displaystyle \Large \boldsymbol{} 3x(x+6)=-10 \\\\3x^2+18x=-10 \\\\3x^2+18x+10=0 \\\\D=324-120=204 \\\\ x_{1;2}=\frac{18\pm2\sqrt{51} }{6} =\frac{9\pm\sqrt{51} }{3}[/tex]

Answer:

C. 7 units

Step-by-step explanation:

The given parameters are;

The length of the chord of the circle, [tex]\overline{AC}[/tex] = 14 units

The orientation of the radius and the chord = The radius is perpendicular to the chord

We have in ΔAOC, [tex]\overline{AO}[/tex] = [tex]\overline{OC}[/tex] = The radius of the circle

[tex]\overline{OB}[/tex] ≅ [tex]\overline{OB}[/tex]  by reflexive property

The angle at point B = 90° by angle formed by the radius which is perpendiclar to the chord [tex]\overline{AC}[/tex]

ΔAOB and ΔCOB are right triangles (triangles having one 90° angle)

[tex]\overline{AO}[/tex] and [tex]\overline{OC}[/tex] are hypotenuse sides of ΔAOB and ΔCOB respectively and [tex]\overline{OB}[/tex] is a leg to ΔAOB and ΔCOB

Therefore;

ΔAOB ≅ ΔCOB, by Hypotenuse Leg rule of congruency

Therefore;

[tex]\overline{AB}[/tex] ≅ [tex]\overline{BC}[/tex] by Congruent Parts of Congruent Triangles are Congruent, CPCTC

[tex]\overline{AB}[/tex] = [tex]\overline{BC}[/tex] by definition of congruency

[tex]\overline{AC}[/tex] = [tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] by segment addition postulate

∴ [tex]\overline{AC}[/tex] = [tex]\overline{AB}[/tex] + [tex]\overline{BC}[/tex] =  [tex]\overline{AB}[/tex] + [tex]\overline{AB}[/tex] = 2 ×  [tex]\overline{AB}[/tex]

∴  [tex]\overline{AB}[/tex] = [tex]\overline{AC}[/tex]/2

[tex]\overline{AB}[/tex] = 14/2 = 7

[tex]\overline{AB}[/tex] = 7 units.

Answer:

7 units

Step-by-step explanation:

Step-by-step explanation:

LCM of 7 and 5 is 35

now will make the denominator equal,

10+7/35

So, Ram's work = 10/35

Shyam's work = 7/35

so

Ram > Shyam

hope it helps