Which diagram best shows how fraction bars can be used to evaluate One-half divided by one-fourth? A fraction bar labeled 1. Under the 1 are 2 boxes containing one-half. Under the 2 boxes are 4 boxes containing one-fourth. 2 one-fourths are circled. A fraction bar labeled one-fourth. Under the one-fourth are 2 boxes containing one-half. Under the 2 boxes are 4 boxes containing 1. 2 boxes containing 1 are circled. A fraction bar labeled 1. Under the 1 are 4 boxes containing one-fourth. Under the 4 boxes are 8 boxes containing one-half. One box containing one-half is circled. A fraction bar labeled 1. Under the 1 are 4 boxes containing one-fourth. Under the 4 boxes are 2 boxes containing one-half. One box containing one-half is circled.

Answer:

x is 5

Step-by-step explanation:

[tex] \frac{4x - 5}{9x - 5} = \frac{3}{8} \\ \\ 8(4x - 5) = 3(9x - 5) \\ 32x - 40 = 27x - 15 \\ 5x = 25 \\ x = \frac{25}{5} \\ \\ x = 5[/tex]

Step-by-step explanation:

as you can see as i solved above. all you need to do was to rationalize the both equations

9514 1404 393

Answer:

  67.0 square units

Step-by-step explanation:

The formula for the area is ...

  Area = 1/2ab·sin(C)

  Area = (1/2)(14)(28)sin(20°) ≈ 67.036 . . . . square units

The area of the triangle is about 67.0 square units.

Answer:

50%

Step-by-step explanation:

35 is halve of 70 therefore it is 50%

hope it helps u...........

Answer:

W=7 and L=11

Step-by-step explanation:

We have two unknowns so we must create two equations.

First the problem states that  length of a rectangle is 10 yd less than three times the width so: L= 3w-10

Next we are given the area so: L X W = 77

Then solve for the variable algebraically. It is just a system of equations.

3W^2 - 10W - 77 = 0

(3W + 11)(W - 7) = 0

W = -11/3 and/or W=7

Discard the negative solution as the width of the rectangle cannot be less then 0.

So W=7

Plug that into the first equation.

3(7)-10= 11 so L=11

Answer:

0.1088 or 10.88%

Step-by-step explanation:

q = 1 - 0.35 = 0.65

P(X=2) = 12C2 × (0.35)² × (0.65)¹⁰

= 0.1088

Answer:

B. 1 + ln 2 - ln x

General Formulas and Concepts:

Algebra II

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle ln(\frac{2e}{x})[/tex]

Step 2: Simplify

Answer:40, 80 and 62

Step-by-step explanation:

182-22= 160

160/4 = 40 so,

Shortest side is 40

Longest is 80

Third side is 62

Answer:

The answer is "0.07404893".

Step-by-step explanation:

Applying the binomial distribution:

[tex]n = 22\\\\p= 82\%=0.82\\\\q = 1-0.82 = 0.18\\\\[/tex]

Calculating the probability for not enough seats:

[tex]=P(X>20)\\\\= P(21) + P(22)\\\\[/tex]

[tex]= \binom{22}{21} (0.82)^{21}(0.18)^1+ \binom{22}{22} (0.82)^{22}(0.18)[/tex]

[tex]=0 .06134598+ 0.01270295\\\\=0.07404893[/tex]

Volume = πr²h

Radius = 3yd

Height = 12yd

Take π = 22/7

Volume = 22/7×3×3×12

= 2376/7

= 339.4285714yd³

Rounding off to nearest tenth

= 339.43yd³

Answered by Gauthmath must click thanks and mark brainliest

Answer:

distributive

Step-by-step explanation:

a(b + c)=ab + ac

it's distributive one

Answer:

5880 ways

Step-by-step explanation:

For selections like this, we solve using the combination theory. Recall that

nCr = n!/(n-r)!r!

Hence given to find the number of ways 6 gentleman and 4 ladies can be choosen out of 10 gentleman and 8 ladies,

= 10C6 * 8C4

= 10!/(10-6)!6! * 8!/(8-6)!6!

= 10 * 9 * 8 * 7 * 6!/4 *3 *2 * 6! * 8 * 7 * 6!/2 * 6!

= 210 * 28

= 5880 ways

The arrangement can be done in 5880 ways

Answer:

Step-by-step explanation:

Side=AC=9[tex]\sqrt{2}[/tex]

Side AB= x

Hypotenuse =CB= y

Side AB = 9[tex]\sqrt{2}[/tex]

Hypotenuse CB = 36

Answer:

68.3 degrees

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

tan I = opp side / adj side

tan I = sqrt(82) / sqrt(13)

tan I = sqrt(82/13)

Taking the inverse tan of each side

tan ^-1 ( tan I) = tan ^-1( sqrt(82/13))

I = 68.2892

Rounding to the nearest tenth

I = 68.3 degrees

Answer:

Step-by-step explanation:

Answer:

I hope it help and I guess it is correct

Answer:

a) Everyone on the team talks until the entire team agrees on one decision.

Step-by-step explanation:

Option B consists of voting and not everyone would like the outcome. Option C is making an outsider the decision maker, which can't be helpful since he / she won't have as strong opinions as the team itself. Option D is just plain wrong as it defeats the purpose of team work and deciding as one team. So, I believe option A makes the most sense

Answer:

Not a function

Domain: {3,4}

Range: {4,5}

Step-by-step explanation:

A function is a relation where each input has its own output. In other words if the x value has multiple corresponding y values then the relation is not a function

For the relation given {(3, 4), (3, 5), (4, 4), (4, 5)} the x value 3 and 4 have more than one corresponding y value therefore the relation shown is not a function

Now let's find the domain and range.

Domain is the set of x values in a relation.

The x values of the given relation are 3 and 4 so the domain is {3,4}

The range is the set of y values in a relation

The y value of the given relation include 4 and 5

So the range would be {4,5}

Notes:

The values of x and y should be written from least to greatest when writing them out as domain and range.

They should be written inside of brackets

Do not repeat numbers when writing the domain and range

Answer:

0.0022 = 0.22% probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other.

Step-by-step explanation:

The trees are arranged, so the arrangements formula is used to solve this question. Also, a probability is the number of desired outcomes divided by the number of total outcomes.

Arrangements formula:

The number of possible arrangements of n elements is given by:

[tex]A_n = n![/tex]

Desired outcomes:

Two cases:

6 redwoods(6! ways) then the 6 pine trees(6! ways)

6 pine trees(6! ways) then the 6 redwoods(6! ways)

So

[tex]D = 2*6!*6![/tex]

Total outcomes:

12 trees, so:

[tex]D = 12![/tex]

What is the probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other?

[tex]p = \frac{D}{T} = \frac{2*6!*6!}{12!} = 0.0022[/tex]

0.0022 = 0.22% probability that he randomly plants the trees so that all 6 redwoods are next to each other and all 6 pine trees are next to each other.

Answer:

108.82

Step-by-step explanation:

Answer:

0.0022 = 0.22% probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

In this question, the elements are arranged, so we have to use the arrangements formula.

Arrangements formula:

The number of possible arrangements of n elements is:

[tex]A_{n} = n![/tex]

Desired outcomes:

Pine trees(6!) then the willows(6!) or

Willows(6!) then the pine trees(6!). So

[tex]D = 2*6!*6! = 1036800 [/tex]

Total outcomes:

12 trees, so:

[tex]T = 12! = 479001600 [/tex]

What is the probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other?

[tex]p = \frac{D}{T} = \frac{1036800 }{479001600 } = 0.0022[/tex]

0.0022 = 0.22% probability that she randomly plants the trees so that all 6 pine trees are next to each other and all 6 willows are next to each other.

Answer:

a) 0.0081 = 0.81% probability that he receives no job offer

b) He expects to get 2.8 job offers.

c) 0.0837 = 8.37% probability that more than half of the firms he applied do not make him any offer.

d) Each job must be independent of other jobs. Additionaly, if [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the normal approximation to the binomial distribution can be used.

e) 0.2401 = 24.01% probability of him securing more than 3 offers.

Step-by-step explanation:

For each application, there are only two possible outcomes. Either he gets an offer, or he does not. The probability of getting an offer for a job is independent of any other job, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

He can expect to receive a job offer from 70% of the firms to which he applies.

This means that [tex]p = 0.7[/tex]

The student decides to apply to only four firms.

This means that [tex]n = 4[/tex]

(a) What is the probability that he receives no job offer?

This is [tex]P(X = 0)[/tex]. So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{4,0}.(0.7)^{0}.(0.3)^{4} = 0.0081[/tex]

0.0081 = 0.81% probability that he receives no job offer.

(b) How many job offers he expects to get?

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

In this question:

[tex]E(X) = 4(0.7) = 2.8[/tex]

He expects to get 2.8 job offers.

(c) What is the probability that more than half of the firms he applied do not make him any offer?

Less than 2 offers, which is:

[tex]P(X < 2) = P(X = 0) + P(X = 1)[/tex]

So

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{4,0}.(0.7)^{0}.(0.3)^{4} = 0.0081[/tex]

[tex]P(X = 1) = C_{4,1}.(0.7)^{1}.(0.3)^{3} = 0.0756[/tex]

Then

[tex]P(X < 2) = P(X = 0) + P(X = 1) = 0.0081 + 0.0756 = 0.0837[/tex]

0.0837 = 8.37% probability that more than half of the firms he applied do not make him any offer.

(d) What assumptions do you need to make to find the probabilities? To increase the chance of securing more job offers, the student decides to apply to as many companies as possible, he sent out 60 applications to all different accounting firms.

Each job must be independent of other jobs. Additionaly, if [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], the normal approximation to the binomial distribution can be used.

(e) What is the probability of him securing more than 3 offers?

Between 4 and n, since n is 4, 4 offers, so:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 4) = C_{4,4}.(0.7)^{4}.(0.3)^{0} = 0.2401[/tex]

0.2401 = 24.01% probability of him securing more than 3 offers.

Answer:

[tex]56[/tex] choices

Step-by-step explanation:

We know that we'll have to solve this problem with a permutation or a combination, but which one do we use? The answer is a combination because the order in which the child picks the candy does not matter.

To further demonstrate this, imagine I have 4 pieces of candy labeled A, B, C, and D. I could choose A, then C, then B or I could choose C, then B, then A, but in the end, I still have the same pieces, regardless of what order I pick them in. I hope that helps to understand why this problem will be solved with a combination.

Anyways, back to the solving! Remember that the combination formula is

[tex]_nC_r=\frac{n!}{r!(n-r)!}[/tex], where n is the number of objects in the sample (the number of objects you choose from) and r is the number of objects that are to be chosen.

In this case, [tex]n=8[/tex] and [tex]r=3[/tex]. Substituting these values into the formula gives us:

[tex]_8C_3=\frac{8!}{3!5!}[/tex]

[tex]= \frac{8*7*6*5*4*3*2*1}{3*2*1*5*4*3*2*1}[/tex] (Expand the factorials)

[tex]=\frac{8*7*6}{3*2*1}[/tex] (Cancel out [tex]5*4*3*2*1[/tex])

[tex]=\frac{8*7*6}{6}[/tex] (Evaluate denominator)

[tex]=8*7[/tex] (Cancel out [tex]6[/tex])

[tex]=56[/tex]

Therefore, the child has [tex]\bf56[/tex] different ways to pick the candies. Hope this helps!

Base case (n = 1):

• left side = 1×2² = 4

• right side = 1×(1 + 1)×(3×1² + 11×1 + 10)/12 = 4

Induction hypothesis: Assume equality holds for n = k, so that

1×2² + 2×3² + 3×4² + … + k × (k + 1)² = k × (k + 1) × (3k ² + 11k + 10)/12

Induction step (n = k + 1):

1×2² + 2×3² + 3×4² + … + k × (k + 1)² + (k + 1) × (k + 2)²

= k × (k + 1) × (3k ² + 11k + 10)/12 + (k + 1) × (k + 2)²

= (k + 1)/12 × (k × (3k ² + 11k + 10) + 12 × (k + 2)²)

= (k + 1)/12 × ((3k ³ + 11k ² + 10k) + 12 × (k ² + 4k + 4))

= (k + 1)/12 × (3k ³ + 23k ² + 58k + 48)

= (k + 1)/12 × (3k ³ + 23k ² + 58k + 48)

On the right side, we want to end up with

(k + 1) × (k + 2) × (3 (k + 1) ² + 11 (k + 1) + 10)/12

which suggests that k + 2 should be factor of the cubic. Indeed, we have

3k ³ + 23k ² + 58k + 48 = (k + 2) (3k ² + 17k + 24)

and we can rewrite the remaining quadratic as

3k ² + 17k + 24 = 3 (k + 1)² + 11 (k + 1) + 10

so we would arrive at the desired conclusion.

To see how the above rewriting is possible, we want to find coefficients a, b, and c such that

3k ² + 17k + 24 = a (k + 1)² + b (k + 1) + c

Expand the right side and collect like powers of k :

3k ² + 17k + 24 = ak ² + (2a + b) k + a + b + c

==>   a = 3   and   2a + b = 17   and   a + b + c = 24

==>   a = 3, b = 11, c = 10

========================================================

Explanation:

At 95% confidence, the z critical value is roughly z = 1.960

The population standard deviation is given to be sigma = 3.7

The error is E = 2 since we want to be within 2 inches of the population mean mu

The min sample size needed is:

n = (z*sigma/E)^2

n = (1.960*3.7/2)^2

n = 13.147876

n = 14

We always round up to the nearest whole number to ensure that we clear the hurdle (otherwise, the sample is too small). It doesn't matter that we're closer to 13 than to 14.

Given:

The given expression is:

[tex]6x^2y-3xy-24xy^2+12y^2[/tex]

To find:

Part A: The expression by factoring out the greatest common factor.

Part B: Factor the entire expression completely.

Solution:

Part A:

We have,

[tex]6x^2y-3xy-24xy^2+12y^2[/tex]

Taking out the highest common factor 3y, we get

[tex]=3y(2x^2-x-8xy+4y)[/tex]

Therefore, the required expression is [tex]3y(2x^2-x-8xy+4y)[/tex].

Part B:

From part A, we have,

[tex]3y(2x^2-x-8xy+4y)[/tex]

By grouping method, we get

[tex]=3y(x(2x-1)-4y(2x-1))[/tex]

[tex]=3y(x-4y)(2x-1)[/tex]

Therefore, the required factored form of the given expression is [tex]3y(x-4y)(2x-1)[/tex].

Answer:

Step-by-step explanation:

a) categorical

b) add all of the numbers and divide by how many numbers there were.

c) outliers means any that were far away from the rest of the data

d) not entirely, you can make an estimate based on it, but nat an exact answer.

Answer:

BC^2=10^2+30^2

Step-by-step explanation:

Using pythagorean theorem

[tex]\\ \sf\longmapsto BC^2=10^2+30^2[/tex]

[tex]\\ \sf\longmapsto BC^2=100+300[/tex]

[tex]\\ \sf\longmapsto BC^2=400[/tex]

[tex]\\ \sf\longmapsto BC=\sqrt{400}[/tex]

[tex]\\ \sf\longmapsto BC=20[/tex]