There are 20 children music class. If 6 children can sit at a table how many
tables de they need?

Answer:

he first one worked 15 hours, the second one worked 5 hours.

Step-by-step explanation:

First mechanic worked x hours, then second one worked 20 - x hours

First one charged $95*x, second one charged $75*(20-x)

Total charge = 95x + 75(20 - x) = 1800

95x + 1500 - 75x = 1800

20x = 300

x = 15, 20 - x = 5

So the first one worked 15 hours, the second one worked 5 hours.

Answer:

The juice bar can sell 3 more cups of the passion juice than the berry juice (last one)

Step-by-step explanation:

Passion: 768/16=48

Berry: 810/18=45

passion can fill 48 cups while berry can fill 45 cups, so passion fills +3

:)

Answer:1244

Step-by-step explanation:

Answer:

Step-by-step explanation:

Use the formula for calculating compound interest PN=P0(1+rk)Nk where N is the unknown, PN=2250, k=1, N=15, and r=0.028. Substitute the values into the formula and simplify.

2250=P(1+0.0281)1⋅15

2250=P(1.028)15

2250=P(1.5132...)

1486.91=P

Rounded to the nearest dollar, Lucy's initial deposit was $1487.

Answer:

[tex]y = \frac{x}{2} + 4[/tex]

Answer:

77

Step-by-step explanation:

<DBE and <ABE are supplementary angles so their sum is equal to 180

<DBE is given as 103

103 + <ABE = 180 subtract 103 from both sides

<ABE = 77

Answer:

cable 1 ≈ 54.63 ft

Step-by-step explanation:

using Pythagoras' identity in the right triangle having cable 1 as its hypotenuse.

let cable 1 be x , then

x² = 22² + 50² = 484 + 2500 = 2984 ( take square root of both sides )

x = [tex]\sqrt{2984}[/tex] ≈ 54.63 ft ( to 2 dec. places )

As rest angles have sum 150°

Using angle sum property

Now

Complementary angles have sum 90°

So the complementary angle is 90-30=60°

Option B

Answer:

-1/4x+3/4

or

-0.25x+0.75

Step-by-step explanation:

g(x) = mx + b

m = slope

b = initial value

m = -1/4

b = 3/4

g(x) = -1/4x+3/4

Hope this helped, let me know if you need more help!

Answer:

28.26

Step-by-step explanation:

[tex]3.14\times9[/tex]

Calculate the product or quotient
[tex]28.26[/tex]

I hope this helps you

:)

Answer:

<aob=1/2(can + ab)

<aob=1/2(25+46)

<aob=1/2(71)

<aob=35.5

Answer:48 mm

Step-by-step explanation:

6 x 2 x 4= 48mm

Answer:

y = 90[tex]\sqrt{2}[/tex]

Step-by-step explanation:

using the sine ratio in the right triangle and the exact value

sin45° = [tex]\frac{1}{\sqrt{2} }[/tex] , then using the 45° angle on the lower left

sin45° = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{90}{y}[/tex] = [tex]\frac{1}{\sqrt{2} }[/tex] ( cross- multiply )

y = 90[tex]\sqrt{2}[/tex]

Answer:

  20

Step-by-step explanation:

T19 = 97

T20 = 102

The 20th term is the first term more than 100.

__

You can find the value of n from ...

  2 +5n > 100

  5n > 98

  n > 19.6

The 20th term is the first term more than 100.

Answer:

-x^2 +10x + 7

[tex]-x^2 +10x + 7[/tex]

Step-by-step explanation:

its the gray shaded area [(x+1)*(x+7)] minus the white area [(x-1)*(2x)]

AKA: [(x+1)*(x+7)] - [(x-1)*(2x)]

so first you foil to get both areas

gray region: (x+1)*(x+7) = x^2 + 8x + 7

white region: (x-1)*(2x) = 2x^2 - 2x

and then you subtract the white from the gray

[tex](x^2 + 8x + 7) - (2x^2 - 2x)[/tex]

-x^2 +10x + 7

Answer:

Step-by-step explanation:

Given data

total distance=160 cm

time taken =20seconds

We know that speed =distance/time

So let us solve for the speed of the ant

speed=160/20

speed= 8cm/s

Also when the distance is 216,and the ant uses the same speed,it will take the ant

time=distance/speed

time=216/8

time= 27seconds

Answer:

1

Step-by-step explanation:

Anything to the exponent of 0 is 1.

Answer:

?

Step-by-step explanation:

Answer:

[tex]\sf x= 14.04^{\circ \:}[/tex]

using tan rule:

[tex]\sf tan(x)= \dfrac{opposite}{adjacent}[/tex]

solve:

[tex]\rightarrow \sf tan(x)= \dfrac{5}{20}[/tex]

[tex]\rightarrow \sf x= tan^{-1}(\dfrac{5}{20})[/tex]

[tex]\rightarrow \sf x= 14.04^{\circ \:}[/tex]

Answer:

x = 14.04° (nearest hundredth)

Step-by-step explanation:

Use the tan trig ratio:

[tex]\mathsf{\tan(\theta)=\dfrac{O}{A}}[/tex]

where:

Given:

[tex]\implies \mathsf{\tan(x)=\dfrac{5}{20}}[/tex]

[tex]\implies \mathsf{x=\arctan\left(\dfrac{5}{20}\right)}[/tex]

[tex]\implies \mathsf{x=14.04\textdegree \ (nearest \ hundredth)}[/tex]

Answer:

2/5x6/7 = 12/35

Step-by-step explanation:

when we multiply fractions we multiply the denominators together and the numerators as well, so 2 x 6 = 12. and  5 x 7 =35

12/35

Answer: 72 square feet

Step-by-step explanation: just multiply 8* 7 = 56 then 8*2 = 16 and add = 72

Answer:

a reflection over the vertical axis, the coordinates would be (-2,5) the first number being the horizontal, second vertical

Answer:

i think its a

Step-by-step explanation:

Based on the estimate from the Waldfogel study, 10[tex]\%[/tex] of this would be a deadweight loss, i.e., [tex]\[/tex]90.6 would be the deadweight loss.

Deadweight loss means the economic inefficiency that takes place when the allotment of resources in a market isn't at the optimal position.

The study suggested that the average recipient's valuation of the gift received was approximately 90[tex]\%[/tex] of the actual purchase price of the gift.

This means that 10[tex]\%[/tex] of the value of the gift gets lost in between.

This loss of 10[tex]\%[/tex] in value constitutes the deadweight loss.

The average amount spent on gifts = $906

Percentage loss in value = 10[tex]\%[/tex] or 0.10

Calculation of the deadweight loss,

DWL = Average amount spent on gifts × Percentage loss in value

DWL = [tex]\[/tex]906 × 0.10

Thus, the deadweight loss would be [tex]\[/tex]90.6.

Learn more about a deadweight loss here:

https://brainly.com/question/32082652

#SPJ12

Answer:

meter (m) is the standard unit measurement for length .

Answer:

it is meter just to let yall know

Step-by-step explanation:

Answer:

£0.60

Step-by-step explanation:

If each pack costs £1.59 and Nadia orders 15 packs,

then the total order before discount = 1.59 x 15 = £23.85

From the table given, we can see that for an order of £23.85 a 2.5% discount will be applied.

Divide £23.85 by 100 to get 1%:  £23.85 ÷ 100 = £0.2385

Now multiply by 2.5 to get 2.5%:  £0.2385 × 2.5 = £0.59625 = £0.60

Alternatively, the calculation in one expression is:

(1.59 × 15) × 0.025 = 0.59625

Complete options are;

a. The approximation requires np > 10 and n(1 - p) > 10.

b. The sample size here is too small to use the Normal approximation to the binomial.

c. The approximation requires np > 30.

d. The Normal approximation works better if the success probability p is close to p = 0.5.

Answer:

Option C is false

Step-by-step explanation:

Looking at the options,

In normal approximation to the binomial,

n is the sample size,

p is the given probability.

q = 1 - p

Now, one of the conditions for using normal approximation to the binomial is that; np and nq or n(1 - p) must be greater than 10.

This means that option A is true because we require np or n(1 - p) to be greater than 10.

From Central limit theorem, the sample size needs to be more than 30 for us to use normal approximation. Our sample is 10. Thus, option B is true.

The approximation doesn't require np > 30. Rather it's the sample size that needs to be more than 30. Thus, option C is false.

Generally, when the value of p in a binomial is close to 0.5, the normal approximations will work better than when the value of p is closer to either 0 or 1. The reason is that: for p = 0.5, the binomial distribution will be symmetrical. Thus, option D is correct.

First look for the fundamental solutions by solving the homogeneous version of the ODE:

[tex]y''+3y'=0[/tex]

The characteristic equation is

[tex]r^2+3r=r(r+3)=0[/tex]

with roots [tex]r=0[/tex] and [tex]r=-3[/tex], giving the two solutions [tex]C_1[/tex] and [tex]C_2e^{-3t}[/tex].

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

[tex]y''+3y'=2t^4[/tex]

Assume the ansatz solution,

[tex]{y_p}=at^5+bt^4+ct^3+dt^2+et[/tex]

[tex]\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e[/tex]

[tex]\implies {y_p}''=20at^3+12bt^2+6ct+2d[/tex]

(You could include a constant term f here, but it would get absorbed by the first solution [tex]C_1[/tex] anyway.)

Substitute these into the ODE:

[tex](20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4[/tex]

[tex]15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4[/tex]

[tex]\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}[/tex]

[tex]y''+3y'=t^2e^{-3t}[/tex]

[tex]e^{-3t}[/tex] is already accounted for, so assume an ansatz of the form

[tex]y_p=(at^3+bt^2+ct)e^{-3t}[/tex]

[tex]\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}[/tex]

[tex]\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}[/tex]

Substitute into the ODE:

[tex](9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}[/tex]

[tex]9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2[/tex]

[tex]-9at^2+(6a-6b)t+2b-3c=t^2[/tex]

[tex]\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}[/tex]

[tex]y''+3y'=\sin(3t)[/tex]

Assume an ansatz solution

[tex]y_p=a\sin(3t)+b\cos(3t)[/tex]

[tex]\implies {y_p}'=3a\cos(3t)-3b\sin(3t)[/tex]

[tex]\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)[/tex]

Substitute into the ODE:

[tex](-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)[/tex]

[tex](-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)[/tex]

[tex]\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}[/tex]

So, the general solution of the original ODE is

[tex]y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}[/tex]