537 miles in 30 hours

Answer:

0.872

Step-by-step explanation:

Given that :

P(cloudy) = P(C) = 0.94

P(cloudy and rainy) = P(C n R) = 0.82

Probability that a given day will be rainy if it is cloudy ; this is a conditional probability problem:

Recall ; P(A|B) = P(AnB) / P(B)

P(R|C) = P(C n R) / P(C) = 0.82 / 0.94 = 0.872

Answer:

Step-by-step explanation: She has 2 more nickels then dimes not 2 times more therefore answers B and D are incorrect. C is incorrect because it has that there are 2 more dimes than nickels. A is correct because it says that there are c dimes, and then c +2 nickels.

Step-by-step explanation:

area of a circle is r x r x pi

so one quarter of it us r x r x pi /4

Answer:

I think the answer for this is the Dimm slots

Answer:

The value of h(g(3)) is 17.

Step-by-step explanation:

We are given these following functions:

[tex]g(x) = x^2 - 5[/tex]

[tex]h(x) = 7x - 11[/tex]

h(g(3)) ?

[tex]h(g(x)) = h(x^2-5) = 7(x^2-5) - 11 = 7x^2 - 35 - 11 = 7x^2 - 46[/tex]

At x = 3

[tex]h(g(3)) = 7(3)^2 - 46 = 63 - 46 = 17[/tex].

The value of h(g(3)) is 17.

Answer:

Well, 10% of 6640 is $664, and $332 is half of that, so 5%

Missing: ​$9850 ‎​$591.

Step-by-step explanation:

Answer:

espero ayudarte ..............

Answer:

I think B

Step-by-step explanation:

121.346° is more close to 121.3°, than 121.4°

if i'm wrong, the i'm sorry

Answer:

[tex]{ \tt{3 \frac{4}{9} \div (5 \frac{1}{3} - 2 \frac{3}{4}) + 5 \frac{9}{10} }} \\ \\ = { \tt{ \frac{31}{9} \div ( \frac{16}{3} - \frac{11}{4} ) + \frac{59}{10} }} \\ \\ = { \tt{ \frac{31}{9} \div ( \frac{31}{12} ) + \frac{59}{10} }} \\ \\ { \tt{ = \frac{4}{3} + \frac{59}{10} }} \\ \\ { \bf{ = \frac{217}{30} }} \\ \\ { \boxed{ \tt{answer : 7 \frac{7}{30} }}} \\ \\ { \underline{ \blue{ \tt{becker ⚜jnr}}}}[/tex]

Answer:

[tex]7 \frac{7}{30}[/tex]

Step-by-step explanation:

[tex]3 \frac{4}{9} \div ( 5\frac{1}{3} - 2 \frac{3}{4}) + 5 \frac{9}{10}\\\\\frac{31}{9} \div (\frac{16}{3} - \frac{11}{4} ) + \frac{59}{10} \\\\\\Solving \ using \ BODMAS\\\\First \ Solve \ expression \ inside \ Bracket \\\\\frac{31}{9} \div (\frac{(16 \times 4) - ( 11 \times 3)}{12}) + \frac{59}{10} \\\\\frac{31}{9} \div (\frac{64- 33)}{12}) + \frac{59}{10} \\\\\frac{31}{9} \div \frac{31}{12} + \frac{59}{10} \\\\\\ \\\\\\Next \ solve \ Dvision \\\\\frac{\frac{31}{9}}{\frac{31}{12}} + \frac{59}{10}\\\\[/tex]

[tex](\frac{31}{9}} \times {\frac{12}{31}) + \frac{59}{10}[/tex]

[tex]\frac{4}{3} + \frac{59}{10}\\\\ Now \ solve \ final \ expression \\\\\\\frac{(4 \times 10) + ( 59 \times 3)}{30}\\\\\frac{40 + 177}{30}\\\\\frac{217}{30}\\\\7 \frac{7}{30}[/tex]

multiplications

mark me brainliesttt :))

Answer:

The work done is 202.50Nm

Step-by-step explanation:

Given

[tex]F =450N[/tex]

[tex]x_1 = 30cm[/tex]

[tex]x_2 = 60cm[/tex]

Required

The work done

First, we calculate the spring constant (k)

[tex]F = kx_1[/tex]

[tex]450N = k *30cm[/tex]

[tex]k = \frac{450N}{30cm}[/tex]

[tex]k =15N/cm[/tex]

So:

[tex]F = kx_1[/tex]

[tex]F(x) = 15x[/tex]

The work done using Hooke's law is:

[tex]W =\int\limits^a_b {F(x)} \, dx[/tex]

This gives:

[tex]W =\int\limits^{60}_{30} {15x} \, dx[/tex]

Rewrite as:

[tex]W =15\int\limits^{60}_{30} {x} \, dx[/tex]

Integrate

[tex]W =15 \frac{x^2}{2}|\limits^{60}_{30}[/tex]

This gives:

[tex]W =15 *\frac{60^2 - 30^2}{2}[/tex]

[tex]W =15 *\frac{2700}{2}[/tex]

[tex]W =15 *1350[/tex]

[tex]W =20250N-cm[/tex]

Convert to Nm

[tex]W =\frac{20250Nm}{100}[/tex]

[tex]W =202.50Nm[/tex]

9514 1404 393

Answer:

 1.60

Step-by-step explanation:

The growth factor is 1 more than the growth rate:

  1 + 60% = 1 + 0.60 = 1.60 = growth factor

Given:

The function is

[tex]f(x)=(x-5)^2(3-x)^2[/tex]

To find:

The derivative of the given function.

Solution:

Chain rule of differentiation:

[tex][f(g(x))]'=f'(g(x))g'(x)[/tex]

Product rule of differentiation:

[tex][f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)[/tex]

We have,

[tex]f(x)=(x-5)^2(3-x)^2[/tex]

Differentiate with respect to x.

[tex]f'(x)=(x-5)^2\dfrac{d}{dx}(3-x)^2+(3-x)^2\dfrac{d}{dx}(x-5)^2[/tex]

[tex]f'(x)=(x-5)^2[2(3-x)(0-1)]+(3-x)^2[2(x-5)(1-0)][/tex]

[tex]f'(x)=(x^2-10x+25)(-6+2x)+(9-6x+x^2)(2x-10)[/tex]

[tex]f'(x)=(x^2)(-6)+(-10x)(-6)+(25)(-6)+(x^2)(2x)-10x(2x)+25(2x)+(9)(2x)+(-6x)(2x)+x^2(2x)+9(-10)+(-6x)(-10)+x^2(-10)[/tex]

On further simplification, we get

[tex]f'(x)=-6x^2+60x-150+2x^3-20x^2+50x+18x-12x^2+2x^3-90+60x-10x^2[/tex]

[tex]f'(x)=(2x^3+2x^3)+(-6x^2-20x^2-12x^2-10x^2)+(60x+50x+18x+60x)+(-90-150)[/tex]

[tex]f'(x)=4x^3-48x^2+188x-240[/tex]

Therefore, the derivative of the given function is [tex]f'(x)=4x^3-48x^2+188x-240[/tex].

Answer:

Arc GH = 78°

Step-by-step explanation:

Inscribed angle = m<GIH = 39°

Measure of arc related to inscribed angle = arc GH = ?

Thus:

m<GIH = ½(arc GH) => Inscribed angles theorem

Substitute

39° = ½(arc GH)

Multiply both sides by 2

2*39° = arc GH

78° = arc GH

Arc GH = 78°

Answer:

        8

        8

        8

   8   8

8  8  8

-------------------

1   0   0  0

Step-by-step explanation:

Answer:

0.64(x) = 30

Step-by-step explanation:

Hope that's correct.

Answer:

x = 3 + 2y

Step-by-step explanation:

x - 2y = 3 ( Isolate x on the left side by adding 2y to both sides )

x = 3 + 2y

Step-by-step explanation:

Step 1: Add -y^3 to both sides.

y3+x2+−y3=1+−y3

x2=−y3+1

Step 2: Take square root.

x=√−y3+1 or x=−√−y3+1

Answer:

x=√−y3+1  or  x=−√−y3+1

(I think this is right) (tell me if im right plz and thx)

Answer:

68

Step-by-step explanation:

∠ACB = ∠DCE

          = 180-92/2

          =44

∠CDE = 180 - 44 / 2

         68

Answer:

1) Sam only interviewed five people out of his entire ELA class, so he doesn't have a very wide demographic to study

2) All of these people were Sam's friends, so they might all have similar interests or only enjoy certain things, and that doesn't allow for a wide range of data collection

Answer:

y = 1.19x

Step-by-step explanation:

y is the dependent variable (total cost)

x is the independent variable (number of pounds)

Two formulas:

Compound interest - A = P(1 + r/n)^nt

Simple interest = A = P(1 + rt)

A - final amount

P - initial principal balance

r - interest rate

n - number of times interest applied per time period

t - number of time periods elapsed

Suri --

A = 4,000(1+0.03/1)^5

A = $4,637.09

Carter --

A = 4,000(1+0.03*5)

A = $4,600

In 5 years, Suri makes $37.09 more than Carter

Answer:

0 + 0.5, 0 + 0.6 + 0.05, 2 + 0.3 + 0.05, 1 + 0.06

Step-by-step explanation:

expanded form is just taking the number apart from ones to tenths to hundreds and so on, or from ones to tens to hundreds and then so on. then you just split it and put the numbers in added form.

Answer:

A. 11

Step-by-step explanation:

EQ is half of EG

so 22/2 = 11

Step-by-step explanation:

Since the two triangles are similar

Then the ratio of sides are equal

Then MK:ML =JH:JI

Sub in this and you will get x =64.1

Answer:

x = 14

Step-by-step explanation:

Please note, the word trapezium is a synonym for the word trapezoid.

This problem gives one the area of the trapezoid, a well as one of the measurements of a base and the height of the figure. One is asked to find the length of the other base. This can be done by using the formula to find the area of a trapezoid. This formula is the following,

[tex]A=(h)(\frac{b_1+b_2}{2})[/tex]

Where (A) represents the area of a trapezoid, ([tex]b_1[/tex]) and ([tex]b_2[/tex]) represents the bases and (h) represents the height. Substitute in the given values and solve for the unknown base.

[tex]b_1=7\\h=6\\A=84[/tex]

[tex]A=(h)(\frac{b_1+b_2}{2})\\[/tex]

Substitute,

[tex]84=6(\frac{7+b_2}{2})\\[/tex]

Inverse operations,

[tex]84=6(\frac{7+b_2}{2})[/tex]

[tex]14=\frac{7+b_2}{2}[/tex]

[tex]28=7+b_2[/tex]

[tex]14=b_2[/tex]

Answer:

0.48780487804

Answer:

133 fishes

Step-by-step explanation:

Units of food A = 400 units

Units of food B = 400 units

Fish Bass required 2 units of A and 4 units of B.

Fish Trout requires 5 units of A and 2 units of B.

i. For food A,

total units of food A required = 2 + 5

                                                 = 7 units

number of bass and trout that would consume food A = 2 x [tex]\frac{400}{7}[/tex]

                                               = 114.3

number of bass and trout that would consume food A = 114

ii. For food B,

total units of food B required = 4 + 2

                                                = 6 units

number of bass and trout that would consume food B = 2 x [tex]\frac{400}{6}[/tex]

                                                 = 133.3

number of bass and trout that would consume food B = 133

Thus, the maximum number of fish that the lake can support is 133.

Answer:

x = 6 , y = 7

Step-by-step explanation:

solving  by substitution method

x + y = 13

x = 13 - y                          equation (i)

2x - y = 5

substitute the value of x

2(13 - y) - y = 5

26 - 2y - y = 5

26 - 3y = 5

26 - 5 = 3y

21/3 = y

7 = y

substitute the value of y in equation (i)

x = 13 - y

x = 13 - 7

x = 6