what should be add to 7 3/5 to get 18

Answer:

3 and 8 are the extremes of the given proportion (first listed answer option)

Step-by-step explanation:

In a proportion of the type:

[tex]\frac{A}{B} =\frac{C}{D}[/tex]

the values "A" and "D" are called "extremes of the proportion, while "B" and "C" are called the "means" of the proportion.

This comes from writing it as:   A : B = C : D

where we observe that A and D are the more "extreme" values being further away from the equal sign than the quantities B and C which are more towards the middle (center of the proportion).

So in your case, the numbers 3 and 8 are the extremes

Answer:

3 and 8

Step-by-step explanation:

Answer:

Step-by-step explanation:

=

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Answer:

explanation is below happy to help ya!

Step-by-step explanation:

Answer:

A, C, D

Step-by-step explanation:

Answer: A , C , D

Step-by-step explanation: edge2022

Answer:

139.316148[tex]\pi[/tex]cubed inches

Step-by-step explanation:

If needed, use a calculator. Not sure if there's rounding for the final answer, so I will give you the precise answer.

1. Find the radius of the ball. If you know the circumference is 9.42 inches, divided it by 2, which the quotient is 4.71.

2. Remember the formula is V= [tex]\frac{4}{3} \pi r^3[/tex]

3. [tex]\frac{4}{3} \pi (4.71)^3[/tex]

4. [tex]\frac{4}{3} \pi (104.487111^3)[/tex]

5. [tex]4\pi (34.829037)[/tex]

6. 139.316148[tex]\pi[/tex]cubed inches

Answer:

The committes can be selected in [tex]1.752507297 \times 10^{19}[/tex] ways

Step-by-step explanation:

The order in which the members are chosen to the committee is not important. So we use the combinations formula to solve this question.

Combinations formula:

[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]

The two committees must be disjoint.

This means that a person cannot be part of both committes.

If there are 385 members of congress, how many ways could the committees be selected?

Since the committes are disjoint, 5(math) + 5(computer science) = 10 people will be chosen from the set of 385. So

[tex]C_{385,10} = \frac{385!}{10!(385-10)!} = 1.752507297 \times 10^{19}[/tex]

The committes can be selected in [tex]1.752507297 \times 10^{19}[/tex] ways

Answer:

a) The third quartile Q₃ = 56.45

b) The variance = 2633.31

Step-by-step explanation:

a) The coefficient of skewness formula is given as follows;

[tex]SK = \dfrac{Q_{3}+Q_{1}-2Q_{_{2}}}{Q_{3}-Q_{1}}[/tex]

Plugging in the values, we have;

[tex]-0.38 = \dfrac{Q_{3}+30.8-2 \times 48.5_{_{}}}{Q_{3}-30.8}[/tex]

Solving gives Q₃ = 56.45

b) To determine the variance, we use the skewness formula as follows;

[tex]SK_{p} = \dfrac{Mean-\left (3\times Median - 2\times Mean \right )}{\sigma } = \dfrac{3\times\left ( Mean - Median \right )}{\sigma }[/tex]

Plugging in the values, we get;

[tex]-0.38= \dfrac{42-\left (3\times 48.5- 2\times 42\right )}{\sigma } = \dfrac{-19.5}{\sigma}[/tex]

[tex]\therefore \sigma =\dfrac{-19.5}{-0.38} = 51.32[/tex]

The variance = σ² = 51.32² = 2633.31.

9514 1404 393

Answer:

Step-by-step explanation:

If order doesn't matter, the number is the combinations of 50 things taken 20 at a time:

  50C20 = 47129212243960 ≈ 4.713×10^13

If order does matter, then the number of ways is this value multiplied by 20!:

  50P20 = 114660755112113373922453094400000 ≈ 1.147×10^32

_____

nCk = n!/(k!(n-k)!)

nPk = n!/(n-k)!

Answer:

no 8 is not divisible to 343060

Answer:

P( red, red, red no replacement ) =  2/17

P( 3 cards with same rank, no replacement) =  1/425

Step-by-step explanation:

There are 52 cards, 26 are red

P( red) =  red/total =26/ 52 = 1/2

Now draw the second card without replacement

There are 51 cards, 25 are red

P( red) =  red/total =25/51

Now draw the third card without replacement

There are 50 cards, 24 are red

P( red) =  red/total =24/50 = 12/25

P( red, red, red no replacement ) = 1/2 * 25/51 * 12/25 = 2/17

P ( all have the same rank)

There are 52 cards,  4 of each rank

We don't care what the first rank is, now the second and third draw have to have the same rank as the first

P( rank) = 1/1

Now draw the second card without replacement

There are 51 cards, 3 are left with the same rank

P( same rank) = cards with same rank/total = 3/51

Now draw the third card without replacement

There are 50 cards, 2 are left with the same rank

P( same rank) = cards with same rank/total = 2/50 = 1/25

P( 3 cards with same rank, no replacement) = 1 * 3/51*1/25 = 1/425

Answer:

Step-by-step explanation:

Please use " ^ " for exponentiation:  50 - x^2 = 0.

Solving for x^2, we get:                      x^2 = 50

Taking the square root of both sides:    x = ±√(50), or x = ±7.07

None of the three possible answers here is correct.

Answer:

[tex]d = \dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]

Step-by-step explanation:

Given the equation of velocity w.r.to time 't':

[tex]v=9t-\dfrac{2}{9}+t^2 ...... (1)[/tex]

Formula for Displacement:

[tex]Displacement = \text{velocity} \times \text{time}[/tex]

So, if we find integral of velocity w.r.to time, we will get displacement.

[tex]\Rightarrow \text{Displacement}=\int {v} \, dt[/tex]

[tex]\Rightarrow \int {v} \, dt = \int ({9t-\dfrac{2}{9}+t^2}) \, dt \\\Rightarrow \int{9t} \, dt - \int{\dfrac{2}{9}} \, dt + \int{t^2} \, dt\\\Rightarrow s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3} + C ....... (1)[/tex]

Here, C is constant (because it is indefinite integral)

Formula for integration used:

[tex]1.\ \int({A+B}) \, dx = \int {A} \, dx + \int{B} \, dx \\2.\ \int({A-B}) \, dx = \int {A} \, dx - \int{B} \, dx \\3.\ \int{x^{n} } \, dx = \dfrac{x^{n+1}}{n+1}\\4.\ \int{C } \, dx = Cx\ \{\text{C is a constant}\}[/tex]

Now, it is given that s = 0, when t = 0.

Putting the values in equation (1):

[tex]0=\dfrac{9\times 0^{2} }{2} - \dfrac{2}{9}\times 0 + \dfrac{0^3}{3} + C\\\Rightarrow C = 0[/tex]

So, the equation for displacement becomes:

[tex]s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]

Answer:

6000

Step-by-step explanation:

6 six

60 sixty

600 six hundred

6000 six thousand

//have a great day//

Answer: A

Step-by-step explanation: 5y=x+5 simplified to slope-intercept form is y = 1/5x + 1

The equation 5y = x + 5 is a linear equaton, and the graph A represents the equation 5y = x + 5

The equation is given as:

5y = x + 5

Make x = 0.

So, we have:

5y = 0 + 5

This gives

5y = 5

So, we have:

y = 1

The point is represented as (0,1)

Make y = 0.

So, we have:

5 * 0 = x + 5

This gives

0 = x + 5

So, we have:

x = -5

The point is represented as (-5,0)

This means that the graph of 5y = x + 5 passes through the points (0,1) and (-5,0)

From the list of options, only graph (A) passes through the points (0,1) and (-5,0)

Hence, the graph A represents the equation 5y = x + 5

Read more about linear equations at:

https://brainly.com/question/14323743

#SPJ2

Answer:

Oioj[

Step-by-step explanation:

Step-by-step explanation:

Write the equation to solve it in order to know the exact value of X.

Answer:

66g of polyethylene ointment base!

Answer:

The probability of tossing a tail and then rolling a number greater than 5 is 0.188

Step-by-step explanation:

Independent events:

If two events, A and B, are independent, we have that:

[tex]P(A \cap B) = P(A)*P(B)[/tex]

In this question:

The coin and the die are independent. So

Event A: Tossing a tail.

Event B: Rolling a number greater than 5.

Probability of tossing a tail:

Coin can be heads or tails(2 outcomes), so the probability of a tail is [tex]P(A) = \frac{1}{2} = 0.5[/tex]

Probability of rolling a number greater than 5:

8 numbers(1 through 8), 3 of which(6,7,8) are greater than 5. So the probability of rolling a number greater than 5 is [tex]P(B) = \frac{3}{8} = 0.375[/tex]

Probability of A and B:

[tex]P(A \cap B) = P(A)*P(B) = 0.5*0.375 = 0.188[/tex]

The probability of tossing a tail and then rolling a number greater than 5 is 0.188

Answer:

The company must sell 60 or 70 items to obtain a weekly profit of 200.

Step-by-step explanation:

The profit is the difference between the revenue and the cost of a given task, therefore:

[tex]\text{profit} = R - C\\\text{profit} = 66*x - 0.2*x^2 - (40*x + 640)\\\text{profit} = 66*x - 0.2*x^2 - 40*x - 640\\\text{profit} = - 0.2*x^2 + 26*x - 640[/tex]

To have a profit of 200, we need to sell:

[tex]-0.2*x^2 + 26*x - 640 = 200\\-0.2*x^2 + 26*x -840 = 0\text{ } *\frac{-1}{0.2}\\x^2 -130 + -4200 = 0\\x_{1,2} = \frac{-(-130) \pm \sqrt{(-130)^2 - 4*1*(-4200)}}{2*1}\\x_{1,2} = \frac{130 \pm \sqrt{16900 + 16800}}{2}\\x_{1,2} = \frac{130 \pm \sqrt{100}}{2}\\x_{1,2} = \frac{130 \pm 10}{2}\\x_{1} = \frac{130 + 10}{2} = \frac{140}{2} = 70\\ x_{2} = \frac{130 - 10}{2} = \frac{120}{2} = 60[/tex]

The company must sell 60 or 70 items to obtain a weekly profit of 200.

Answer:

The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute

P(x⁻>96) =0.0359

Step-by-step explanation:

Explanation:-

Given sample size 'n' =10

mean of the Population = 90 words per minute

standard deviation of the Population =10 wpm

we will use formula

                           [tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]

Let X⁻  = 96

                           [tex]Z = \frac{96-90 }{\frac{10}{\sqrt{10} } }[/tex]

                          Z =  1.898

The probability that a random sample of 10 second grade students from     the city results in a mean  reading rate of more than 96 words per minute

[tex]P(X^{-}>x^{-} ) = P(Z > z^{-} )[/tex]

                   = 1- P( Z ≤z⁻)

                   = 1- P(Z<1.898)

                   = 1-(0.5 +A(1.898)

                   = 0.5 - A(1.898)

                   = 0.5 -0.4641 (From Normal table)

                  = 0.0359

Final answer:-

The probability that a random sample of 10 second grade students from  

                = 0.0359

Solution,

Length=5

Volume of cube=(L)^3

= 5*5*5

=125

hope it helps

Good luck on your assignment

Answer:

(a)                             n :      20           50          100         500

P (-200 < X - μ < 200) : 0.2886    0.4444    0.5954    0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.

Assuming that the random variable X follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{20}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{20}})[/tex]

                                           [tex]=P(-0.37<Z<0.37)\\\\=P(Z<0.37)-P(Z<-0.37)\\\\=0.6443-0.3557\\\\=0.2886[/tex]

[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{50}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{50}})[/tex]

                                           [tex]=P(-0.59<Z<0.59)\\\\=P(Z<0.59)-P(Z<-0.59)\\\\=0.7222-0.2778\\\\=0.4444[/tex]

[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{100}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{100}})[/tex]

                                           [tex]=P(-0.83<Z<0.83)\\\\=P(Z<0.83)-P(Z<-0.83)\\\\=0.7977-0.2023\\\\=0.5954[/tex]

[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{500}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{500}})[/tex]

                                           [tex]=P(-1.86<Z<1.86)\\\\=P(Z<1.86)-P(Z<-1.86)\\\\=0.9688-0.0312\\\\=0.9376[/tex]

                                 n :      20           50          100         500

P (-200 < X - μ < 200) : 0.2886    0.4444    0.5954    0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ([tex]\bar x[/tex]) approaches the whole population mean ([tex]\mu_{x}[/tex]).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

Using the normal distribution and the central limit theorem, it is found that:

a)

0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.

0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.

0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.

0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.

b)

b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Hence, the probability of sample of size n having a sample mean within 200 is the p-value of [tex]Z = \frac{200}{s}[/tex] subtracted by the p-value of [tex]Z = -\frac{200}{s}[/tex]

In this problem:

Item a:

For samples of 20, [tex]n = 20[/tex], and thus:

[tex]s = \frac{2400}{\sqrt{20}}[/tex]

Then

[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{20}}} = 0.37[/tex]

0.6443 - 0.3557 = 0.2886.

0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.

For samples of 50, [tex]n = 50[/tex], and thus:

[tex]s = \frac{2400}{\sqrt{50}}[/tex]

Then

[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{50}}} = 0.59[/tex]

0.7224 - 0.2776 = 0.4448.

0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.

For samples of 100, [tex]n = 100[/tex], and thus:

[tex]s = \frac{2400}{\sqrt{100}}[/tex]

Then

[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{100}}} = 0.83[/tex]

0.7967 - 0.2033 = 0.5934

0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.

For samples of 500, [tex]n = 500[/tex], and thus:

[tex]s = \frac{2400}{\sqrt{500}}[/tex]

Then

[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{500}}} = 1.86[/tex]

0.9686 - 0.0314 = 0.9372

0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.

Item b:

A larger sample size reduces the standard error, as [tex]s = \frac{\sigma}{\sqrt{n}}[/tex], that is, the standard error is inversely proportional to the square root of the sample size n, meaning that values are closer to the mean. Thus, the correct option is:

b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

A similar problem is given at https://brainly.com/question/24663213

Answer:

- $0.625

Step-by-step explanation:

To win 3 heads must be obtained, the probability of this is:

p = (1/2) ^ 3 = 0.125

Now, let's review the other scenarios:

HHH -> 0.125

HHT

THH ----> p (2H, 1T) = 3 * 0125 = 0.375

HTH

HTT

TTH ----> p (1H, 2T) = 3 * 0125 = 0.375

THT

TTT -> 0.125

So the waiting value would be:

EV = 2 * 0.125 - 1 * 0.375 - 1 * 0.375 - 1 * 0.125

EV = - 0.625

That is to say that the waiting value is - $ 0.625

The outcome og 1 game cannot be predicted but 100 you loss because the expected value is negative

The waiting value would be "-$0.625", the expected value will be negative so the outcome of game 1 can't be predicted but the 100 games you lose.

According to the question,

The probability will be:

→ [tex]p= (\frac{1}{2} )^3[/tex]

     [tex]= 0.125[/tex]

Now,

HHH:

→ 0.125

THH:

→ [tex]p(2H, 1T) = 3\times 0.125[/tex]

                   [tex]= 0.375[/tex]  

TTH:

→ [tex]p(1H, 2T) = 3\times 0.125[/tex]

                   [tex]= 0.375[/tex]

hence,

The waiting value will be:

→ [tex]EV = 2\times 0.125 -1\times 0.375-1\times -0.375-1\times 0.125[/tex]

         [tex]= -0.625[/tex]

Thus the above response is correct.

Learn more:

https://brainly.com/question/13753898

For A: V = B · h  V = 8 · 14  V = 784 units^3.

For B: V = B · h  V = 9 · 9  V = 729 units^3.

For C: V = B · h  V = 5 · 11  V = 220 units^3.

Answer:

X=1 x=-9

Step-by-step explanation:

(x^2)+8x=9

(x^2)+8x-9=0

(x-1)(x+9)=0

x=1 x=-9

Answer: approximately 5148 feet directly north of point A

This is equivalent to 0.975 miles

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

Explanation:

Check out figure 1 (attached image below) to see how the drawing is set up.

Points A and B are separated by 2 miles, or 10560 feet. Use the conversion factor 1 mile = 5280 feet.

Because sound travels at roughly 1100 feet per second (assuming all of the conditions are right), this means that after x seconds, the sound wave has traveled a total distance of 1100x feet. This is the distance from A to C. Point C is the firework location.

The distance from B to C is 1100(x+6) feet because it takes 6 more seconds for the soundwave to travel from C to B, compared from C to A.

Note that triangle ABC is a right triangle. The 90 degree angle is at point A.

We can use the pythagorean theorem to solve for x

See figure 2 in the attached images to see the steps on solving for x. The numbers get quite big, though in the end x is fairly small. We end up with x = 4.68, which is approximate due to the 1100 figure being approximate.

This means the sound wave takes about 4.68 seconds to go from point C to point A.

With this x value, we can compute the expression 1100x to get

1100*x = 1100*4.68 = 5,148 feet

To convert this to miles, divide by 5280

(5148)/(5280) = 0.975

Therefore,

5148 feet = 0.975 miles

Answer:

As x gets smaller, pointing to negative infinity, the value of f decreses, pointing to negative infinity.

As x gets increases, pointing to positve infinity, the value of f decreses, pointing to negative infinity.

Step-by-step explanation:

To find the end behaviour of a function f(x), we calculate these following limits:

[tex]\lim_{x \to +\infty} f(x)[/tex]

And

[tex]\lim_{x \to -\infty} f(x)[/tex]

In this question:

[tex]f(x) = -4x^{6} + 6x^{2} - 52[/tex]

At negative infinity:

[tex]\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} -4x^{6} + 6x^{2} - 52[/tex]

When the variable points to infinity, we only consider the term with the highest exponent. So

[tex]\lim_{x \to -\infty} -4x^{6} + 6x^{2} - 52 = \lim_{x \to -\infty} -4x^{6} = -4*(-\infty)^{6} = -(\infty) = -\infty[/tex]

So as x gets smaller, pointing to negative infinity, the value of f decreses, pointing to negative infinity.

Positive infinity:

[tex]\lim_{x \to \infty} f(x) = \lim_{x \to \infty} -4x^{6} + 6x^{2} - 52 = \lim_{x \to \infty} -4x^{6} = -4*(\infty)^{6} = -(\infty) = -\infty[/tex]

So as x gets increases, pointing to positve infinity, the value of f decreses, pointing to negative infinity.

f(x) = -4x^6 +6x^2-52

The leading coefficient is negative so the left end of the graph goes down.

f(x) is an even function so both ends of the graph go in the same direction.

Answer:

5c

Step-by-step explanation:

5c is the only value you can take out of both factors

Answer:

For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]

In this question:

[tex]p = 0.061, n = 50[/tex]

So

Mean:

[tex]\mu = p = 0.061[/tex]

Standard deviation:

[tex]s = \sqrt{\frac{0.061*0.939}{50}} = 0.034[/tex]

For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.

Answer: Repeated contrast

Step-by-step explanation:

In the two way ANOVA that was performed above, there were 3 groups of which 30 people (15 men and 15 women) participants who had never learnt to play a musical instrument before were recruited and divided equally.

The ANOVA is of repeated measures. There is within group effect, between group effect and interaction effect. The result also shows significant main effect for the gender and the hours used practicing. Hence, to breakdown the effect of gender, the repeated contrast method will be utilized. The repeated contrast method compares the mean of every level with the succeeding level except the last level.