I don't understand word problems can someone please answer it for me and I need it ASAP.

Answer:

Step-by-step explanation:

A school bag contains 12 pens of which 5 are red and the other are black. 4 pens are selected from the bag.

(a) How many different samples of size 4 pens are possible?

12C4=12!/(4!*8!)=495

(b) How many samples have 3 red pens and 1 black pen?

5C3*7C1

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

=>5C3*7C1=10*7=70

(c) How many samples of size 4 contain at least two red pens?

(5C2*7C2)+(5C3*7C1)+(5C4*7C0)

5C2=5!/(2!*3!)=10

7C2=7!/(2!*5!)=21

5C3=5!/(3!*2!)=10

7C1=7!/(1!*6!)=7

5C4=5!/(4!*1!)=5

7C0=7!/(0!*7!)=1

=>(5C2*7C2)+(5C3*7C1)+(5C4*7C0)=285

(d) How many samples of size 4 contain at most one black pen?

(7C1*5C3)+(7C0*5C4)

7C1=7!/(1!*6!)=7

7C0=7!/(0!*7!)=1

5C3=5!/(3!*2!)=10

5C4=5!/(4!*1!)=5

=>(7C1*5C3)+(7C0*5C4)=(7*10)+(1*5)=75

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

the graph has 12 segments so angle enclosed by each segment is [tex] {2\pi\over 12}=\frac{\pi}6[/tex]

anti-clockwise is taken as positive, so if you want positive q, you need to rotate 8 segments [tex] q=8\frac,{\pi}6=\frac{4\pi}3 [/tex] , and and 8 circles or units so r=8

and for a negative angle, you need to rotate clockwise

Which is 4 segments from the horizontal line. so [tex]q=-\frac{2\pi}3[/tex] and r will be same, 8 units.

[not sure about -r so I won't include it in answer]

Answer:

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Step-by-step explanation:

For the first two cases, ( r, θ ) r would be > 0, where r is the directed distance from the pole, and theta is the directed angle from the positive x - axis.

So when r is positive, we can tell that this point is 8 units from the pole, so r is going to be 8 in either case,

( 8, 240° ) - because r is positive, theta would have to be an angle with which it's terminal side passes through this point. As you can see that would be 2 / 3rd of 90 degrees more than a 180 degree angle,or 60 + 180 = 240 degrees.

( 8, - 120° ) - now theta will be the negative side of 360 - 240, or in other words - 120

Now let's consider the second two cases, where r is < 0. Of course the point will still be 8 units from the pole. Again for r < 0 the point will lay on the ray pointing in the opposite direction of the terminal side of theta.

( - 8, 60° ) - theta will now be 2 / 3rd of 90 degrees, or 60 degrees, for - r. Respectively the remaining degrees will be negative, 360 - 60 = 300, - 300. Thus our second point for - r will be ( - 8, - 300° )

_________________________________

So we have the points ( 8, 240° ), ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). However we only want 3 cases, so we have points ( 8, - 120° ), ( - 8, 60° ), and ( - 8, - 300° ). Let's convert the degrees into radians,

Points : ( 8, - 2π/3 ), ( - 8, π/3 ), ( - 8, - 5π/3 )

Answer:

work pictured and shown

Answer:

Last one

Step-by-step explanation:

● [ ( 3^2 × 5^0) / 4 ]^2

5^0 is 1 since any number that has a null power is equal to 1.

●[ (3^2 ×1 ) / 4 ]^2

● (9/4)^2

● 81 / 16

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First we plot these point on a graph as given in attachment.

From the attachment we can observe that AD || BC || x-axis .

also, AB ||CD, that will make ABCD a parallelogram ,  but to confirm we check the property of parallelogram "diagonals bisect each other" , i.e . "Mid point of both diagonals are equal".

Mid point of AC= [tex](\dfrac{-5+4}{2},\dfrac{2+5}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]

Mid point of BD= [tex](\dfrac{-3+2}{2},\dfrac{5+2}{2})=(\dfrac{-1}{2},\dfrac{7}{2})[/tex]

Thus, Mid point of AC=Mid point of BD

i.e. diagonals bisect each other.

That means ABCD is a parallelogram.

Answer: ABCD is a parallelogram.

Step-by-step explanation:

First, we plot these points on a graph as given in the attachment. From the attachment, we can observe that AD || BC || x-axis. Also, AB ||CD, which will make ABCD a parallelogram, but to confirm, we check the parallelogram property "diagonals bisect each other," i.e., "Midpoint of both diagonals is equal."

The midpoint of AC=. The midpoint of BD=. Thus, the Midpoint of AC=Mid point of BD diagonals bisects each other. That means ABCD is a parallelogram.

Answer:

Change due is 3.50

Step-by-step explanation:

20.00-16.50 is 3.50

Answer: $3.50

Step-by-step explanation:

You subtract 20 from 16.50.

Answer:

n + 1

Step-by-step explanation:

Starting with the integer 'n,' we represent the "next integer" by n + 1.

Answer:

6.02

six point zero two

Step-by-step explanation:

Answer:

602 / 100= 6,02

Step-by-step explanation:

602 to divide 100 = 6,02

Answer:

Solution : Option B

Step-by-Step Explanation:

We have the following system of equations at hand here.

{ x = 5 cot(t), y = - 3csc(t) + 4 }

Now instead of isolating the t from either equation, let's isolate cot(t) and csc(t) --- Step #1,

x = 5 cot(t) ⇒ x - 5 = cot(t),

y = - 3csc(t) + 4 ⇒ y - 4 = - 3csc(t) ⇒ y - 4 / - 3 = csc(t)

Now let's square these two equations. We know that csc²θ - cot²θ = 1, so let's subtract the equations  as well. --- Step #2

( y - 4 / - 3 )² = (csc(t))²

- ( x - 5 / 1 )² = (cot(t))²  

___________________

(y - 4)² / 9 - x² / 25 = 1

And as we are subtracting the two expressions, this is an example of a hyperbola. Therefore your solution is option b.

Step-by-step explanation:

utilise the formula a+(n-1)d

a is the first number while d is common difference

Answer:

22

Step-by-step explanation:

Using the formular, Un = a + (n - 1)d

Where n = 10; a = -23; d = 5

U10 = -23 + (9)* 5

U10 = -23 + 45 = 22

Step-by-step explanation:

Using Sine Rule

[tex] \frac{ \sin(a) }{ |a| } = \frac{ \sin(b) }{ |b| } = \frac{ \sin(c) }{ |c| } [/tex]

[tex] \frac{ \sin(42) }{5} = \frac{ \sin(38) }{a} [/tex]

[tex]a = \frac{5( \sin(38))}{ \sin(42) } [/tex]

[tex]a = 4.6[/tex]

[tex] \frac{ \sin(42) }{5} = \frac{ \sin(100) }{b} [/tex]

[tex]b= \frac{5( \sin(100))}{ \sin(42) } [/tex]

[tex]b = 7.4[/tex]

Answer:

z = 83/( -c+6-t)

Step-by-step explanation:

-cz + 6z = tz + 83

Subtract tz from each side

-cz + 6z -tz= tz-tz + 83

-cz + 6z - tz = 83

Factor out z

z( -c+6-t) = 83

Divide each side by ( -c+6-t)

z( -c+6-t)/( -c+6-t)  = 83/( -c+6-t)

z = 83/( -c+6-t)

Answer:

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Step-by-step explanation:

Given that:

[tex]f(t) = 12 cos (t- \dfrac{\pi}{6})[/tex]

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

[tex]f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}][/tex]

[tex]f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}][/tex]

[tex]f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)[/tex]

[tex]L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ][/tex]

[tex]L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}[/tex]

[tex]L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}[/tex]

[tex]L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ][/tex]

Answer:

see explanation

Step-by-step explanation:

(43)

3[tex]x^{5}[/tex] - 75x³ ← factor out 3x³ from each term

= 3x³(x² - 25) ← this is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , thus

x² - 25 = x² - 5² = (x - 5)(x + 5)

Thus

3[tex]x^{5}[/tex] - 75x³ = 3x³(x - 5)(x + 5)

(44)

81c² + 72c + 16 ← is a perfect square of the form

(ac + b)² = a²c² + 2abc + b²

Compare coefficients of like terms

a² = 81 ⇒ a = [tex]\sqrt{81}[/tex] = 9

b² = 16 ⇒ b = [tex]\sqrt{16}[/tex] = 4

and 2ab = 2 × 9 × 4 = 72

Thus

81c² + 72c + 16 = (9c + 4)²

1. 3x^5 -75x³

=3x³(x²-25)

=3x³(x²-5²)

=3x³(x-5)(x+5)

2. 81c²+72c+16

=81c²+36c+36c+16

=9c(9c+4)+4(9c+4)

=(9c+4)(9c+4)

=(9c+4)²

Answer:

mean=87

median=87

Step-by-step explanation:

mean=sum of test score/number of subject

mean=79+91+93+85+86+88/6

mean=522/6

mean=87

Literal meaning of median is medium.

To find the number which lies in the medium, we must rearrange the number in ascending.

79, 91, 93, 85, 86, 88

79, 85, 86, 88, 91, 93

86+88/2=87

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

Answer:

  [tex]\dfrac{3}{(y+2)^3}[/tex]

Step-by-step explanation:

Maybe you want this written using math symbols. It will be ...

  [tex]\boxed{\dfrac{3}{(y+2)^3}}[/tex]

Complete question is;

A machine used to fill gallon-sized paint cans is regulated so that the amount of paint dispensed has a mean of 128 ounces and a standard deviation of 0.20 ounce. You randomly select 35 cans and carefully measure the contents. The sample mean of the cans is 127.9 ounces. Does the machine need to be? reset? Explain your reasoning.

(yes/no)?, it is (very unlikely/ likely) that you would have randomly sampled 35 cans with a mean equal to 127.9 ?ounces, because it (lies/ does not lie) within the range of a usual? event, namely within (1 standard deviation, 2 standard deviations 3 standard deviations) of the mean of the sample means.

Answer:

Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Step-by-step explanation:

We are given;

Mean: μ = 128

Standard deviation; σ = 0.2

n = 35

Now, formula for standard error of mean is given as;

se = σ/√n

se = 0.2/√35

se = 0.0338

Normally, the range of values should be within 2 standard deviations of mean. In this case, normal range of values will be;

μ ± 2se = 128 ± 0.0338

This gives; 127.9662, 128.0338

So, Yes, we should reset the machine because it is unusual to have a mean equal to 127.9 from a random sample of 35 as the mean of 127.9 doesn't fall within range of a usual event with 2 standard deviations of the mean of the sample means.

Answer:

A) C1 = 0.00187 m = 0.187 cm,  C2 = 0.0062 m = 0.62 cm

B)  A sample of how the graph looks like is attached below ( periodic sine wave )

C) w = [tex]\sqrt[4]{3}[/tex] is when the amplitude of the forced response is maximum

Step-by-step explanation:

Given data :

mass = 5kg

length of spring = 10 cm = 0.1 m

f(t) = 10sin(t) N

viscous force = 2 N

speed of mass = 4 cm/s = 0.04 m/s

initial velocity = 3 cm/s = 0.03 m/s

Formulating initial value problem

y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m

spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m

f(t) = 10sin(t/2) N

using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion

the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)

A) finding the solution of the initial value

attached below is the solution and

B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like

C attached below

Answer:

1/8

Step-by-step explanation:

3/8-1/8-1/8=1/8

Complete Question

On the uploaded image is a similar question that will explain the given question

Answer:

The value of k is  [tex]k = 214285.7[/tex]

The percentage  of the oil that will be cleaned is [tex]x = 80.77\%[/tex]

Step-by-step explanation:

From the question we are told that

   The  cost of cleaning up the spillage is  [tex]C = \frac{ k x }{100 - x }[/tex]  [tex]x \le x \le 100[/tex]

     The  cost of cleaning x =  70% of the oil is  [tex]C = \$500,000[/tex]

Now at  [tex]C = \$500,000[/tex] we have  

       [tex]\$ 500000 = \frac{ k * 70 }{100 - 70 }[/tex]

       [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]\$ 500000 = \frac{ k * 70 }{30 }[/tex]

      [tex]k = 214285.7[/tex]

Now  When  [tex]C = \$900,000[/tex]

       [tex]x = 80.77\%[/tex]

Answer:

csctheta= [tex]\frac{\sqrt{13} }{3}[/tex]

Step-by-step explanation:

answer is provided on top

The value of the [tex]\rm cosec \theta = \frac{\sqrt{13} }{3}[/tex]. Cosec is found as the ratio of the hypotenuse and the perpendicular.

The field of mathematics is concerned with the relationships between triangles' sides and angles, as well as the related functions of any angle

The given data in the problem is;

[tex]\rm cot \theta = \frac{2}{3}[/tex]

The [tex]cot \theta[/tex] is found as;

[tex]\rm cot \theta = \frac{B}{P} \\\\ \rm cot \theta = \frac{2}{3} \\\\ B=2 \\\\ P=3 \\\\[/tex]

From the phythogorous theorem;

[tex]\rm H=\sqrt{P^2+B^2} \\\\ \rm H=\sqrt{2^2+3^2} \\\\ H=\sqrt{13} \\\\[/tex]

The value of the cosec is found as;

[tex]\rm cosec \theta = \frac{H}{P} \\\ \rm cosec \theta = \frac{\sqrt{13} }{3}[/tex]

Hence the value of the [tex]\rm cosec \theta = \frac{\sqrt{13} }{3}[/tex].

To learn more about the trigonometry refer to the link;

https://brainly.com/question/26719838

Answer:

  x = 1.00995066776

  x = 2.52925492433

Step-by-step explanation:

This sort of equation is best solved using a graphing calculator. For that purpose, I like to rewrite the equation as a function whose zeros we're seeking. Here, that becomes ...

  [tex]f(x)=\log{(x)}-\log{(x-1)}^2-2\log{(x-1)}[/tex]

The attached graph shows zeros at

  x = 1.00995066776 and 2.52925492433

_____

Comment on the equation

Note that we have taken the middle term to be the square of the log, rather than the log of a square. For the latter interpretation, see mberisso's answer at https://brainly.com/question/17210068

Comment on the answer refinement

We have used Newton's method iteration to refine the solutions to this equation. The solution near 1.00995 requires the initial guess be very close for that method to work properly. Fortunately, the 1.01 value shown on the graph is sufficient for the purpose.

Find the powers [tex]a=\sqrt{2}+\sqrt{3}[/tex]

$a^{2}=5+2 \sqrt{6}$

$a^{3}=11 \sqrt{2}+9 \sqrt{3}$

The cubic term gives us a clue, we can use a linear combination to eliminate the root 3 term $a^{3}-9 a=2 \sqrt{2}$ Square $\left(a^{3}-9 a\right)^{2}=8$ which gives one solution. Expand we have $a^{6}-18 a^{4}-81 a^{2}=8$ Hence the polynomial $x^{6}-18 x^{4}-81 x^{2}-8$ will have a as a solution.

Note this is not the simplest solution as $x^{6}-18 x^{4}-81 x^{2}-8=\left(x^{2}-8\right)\left(x^{4}-10 x^{2}+1\right)$

so fits with the other answers.

Answer:

[tex]y^3 -6y-6[/tex]

Answer:

The first picture's answer would be (6, 21)

Step-by-step explanation:

You have to find the points on the 8th and the 9th day, and then you would add them together, and then divide by two finding the average, which would be 24 and 18, so when added, you get 42, divided by 2 you get 21. You look on the graph for the point with 21, and you find it is on 6.

Answer:

108.

Step-by-step explanation:

-7, -2, 3, 8 is an arithmetic sequence with a1 (first term) = -7 and common difference (d)  = 5.

The 24th term = a1 + (24 - 1)d

=  -7 + 23 * 5

= -7 + 115

= 108.

Answer:

this? hope it helps ........

Answer:

The answer is area=32pi-64 and the perimeter is 8pi

Step-by-step explanation:

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

The null and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.

Answer:

The upper bound of the confidence interval is 0.34

Step-by-step explanation:

Here in this question, we want to calculate the upper bound of the confidence interval.

We start by calculating the margin of error.

Mathematically, the margin of error = 0.29 -0.24 = 0.05

So to get the upper bound of the confidence interval, we simply add this margin of error to the mean

That would be 0.05 + 0.29 = 0.34

volume of a cone

.

.

.

volume of sphere

.

.

number of spheres that can be made......

.

.

hence a hemisphere can be formed

Answer:

1/3(a+b)

hope it helps :>