samantha wants to sort her greeting cards into boxes of 24 cards each. she has 312 greeting cards. How many boxes will she need?

Answers

Answer 1

312 greeting cards in 24-card boxes require 13 boxes.

312/24=13

Answer 2

Samantha will need 13 boxes to sort her 312 greeting cards.

To determine the number of boxes needed, we divide the total number of greeting cards by the number of cards that can fit into one box. Samantha has 312 greeting cards, and each box can hold 24 cards.

First, we perform the division:

[tex]\[ \frac{312}{24} = 13 \][/tex]

Since we are dealing with whole boxes, we do not need to consider any remainder because Samantha cannot use a fraction of a box. Therefore, Samantha will need exactly 13 boxes to accommodate all 312 greeting cards.


Related Questions

The probability that an appliance is currently being repaired is .5. If an apartment complex has 100 such appliances, what is the probability that at least 60 are currently being repaired? Use the normal approximation to the binomial.

Answers

The probability that no more than [tex]25[/tex] were victims of e-mail fraud is [tex]\fbox{0.0278}[/tex].

Further explanation:

Given:

The probability [tex]p[/tex] that an appliance is currently repaired is [tex]0.5[/tex].

The number of complex [tex]n[/tex] are [tex]100[/tex].

Calculation:

The [tex]\bar{X}[/tex] follow the Binomial distribution can be expressed as,

[tex]\bar{X}\sim \text{Binomial}(n,p)[/tex]

Use the normal approximation for [tex]\bar{X}[/tex] as

[tex]\bar{X}\sim \text{Normal}(np,np(1-p))[/tex]

The mean [tex]\mu[/tex] is [tex]\fbox{np}[/tex]

The standard deviation [tex]\sigma\text{ } \text{is} \text{ } \fbox{\begin{minispace}\\ \sqrt{np(1-p)}\end{minispace}}[/tex]

The value of [tex]\mu[/tex] can  be calculated as,

[tex]\mu=np\\ \mu= 100 \times0.5\\ \mu=50[/tex]

The value of [tex]\sigma[/tex] can be calculated as,

[tex]\sigma=\sqrt{100\times0.5\times(1-0.5)} \\\sigma=\sqrt{100\times0.5\times0.5}\\\sigma=\sqrt{25}\\\sigma={5}[/tex]

By Normal approximation \bar{X} also follow Normal distribution as,

[tex]\bar{X}\sim \text{Normal}(\mu,\sigma^{2} )[/tex]

Substitute [tex]50[/tex] for [tex]\mu[/tex] and [tex]25[/tex] for [tex]\sigma^{2}[/tex]

[tex]\bar{X}\sim\text {Normal}(50,25)[/tex]

The probability that at least [tex]60[/tex] are currently being repaired can  be calculated as,

[tex]\text{Probability}=P\left(\bar{X}>60)\right}\\\text{Probability}=P\left(\frac{{\bar{X}-\mu}}{\sigma}>\frac{{(60-0.5)-50}}{\sqrt{25} }\right)\\\text{Probability}=P\left(Z}>\frac{{59.5-50}}{5}}\right)\\\text{Probability}=P\left(Z}>\frac{9.5}{5}}\right)\\\text{Probability}=P(Z}>1.9})[/tex]

The Normal distribution is symmetric.

Therefore, the probability of greater than [tex]1.9[/tex] is equal to the probability of less than [tex]1.9 [/tex].

[tex]P(Z>1.9})=1-P(Z<1.9)\\P(Z>1.9})=1-0.9722\\P(Z>1.9})=0.0278[/tex]

Hence, the probability that no more than [tex]25[/tex] were victims of e-mail fraud is [tex]\fbox{0.0278}[/tex].

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Answer Details:

Grade: College Statistics

Subject: Mathematics

Chapter: Probability and Statistics

Keywords:

Probability, Statistics, Appliance, Apartment complex, Binomial distribution, Normal distribution, Normal approximation, Central Limit Theorem, Z-table, Mean, Standard deviation, Symmetric.

6. Let's say the countdown for a space shuttle launch has begun. At "T minus 27 hours" (that is, 27 hours before launch), a problem occurs. If the technicians have not fixed the problem by T minus 8 hours, the launch will have to be scratched. How much time do the technicians have to correct the problem?

Answers

Answer: the technicians have 19 hours to correct the problem.

Step-by-step explanation:

The countdown started at "T-minus 27 hours" that means that the clock starts ticking backwards and if an hour passes away the lime left for the launch will be 26 hours and the limit the technicians have is T minus 8 hours so the problem can be solved with a simple substraction 27 hours minus 8 hours equals 19 hours.

g A standard deck of playing cards is shuffled and three people each choose a card. Find the probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement, and if the cards are chosen without replacement. (a) The cards are chosen with replacement. (b) The cards are chosen without replacement.

Answers

Answer:

The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement is 0.03125 .

The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement is 0.03058

Step-by-step explanation:

Total no. of cards = 52

Spade cards = 13 (Black)

Club cards = 13 (Black)

Heart cards = 13 (Red)

Diamond cards = 13(Red)

Total red cards = 26

With replacement case:

Probability of getting spade on first draw = [tex]\frac{13}{52}[/tex]

Now card is replaced

Probability of getting spade on second draw = [tex]\frac{13}{52}[/tex]

Now card is replaced

Probability of getting red card on third draw = [tex]\frac{26}{52}[/tex]

So, The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement =  [tex]\frac{13}{52} \times \frac{13}{52} \times \frac{26}{52}[/tex]

                                                                                        =[tex]\frac{1}{32}[/tex]

                                                                                        =  [tex]0.03125[/tex]

Without replacement case:

Probability of getting spade on first draw = [tex]\frac{13}{52}[/tex]

Remaining cards = 51

Remaining spade cards = 12

Probability of getting spade on second draw = [tex]\frac{12}{51}[/tex]

Remaining cards = 50

Probability of getting red card on third draw = [tex]\frac{26}{50}[/tex]

So, The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement =  [tex]\frac{13}{52} \times \frac{12}{51} \times \frac{26}{50}[/tex]

                                                                                       = [tex]\frac{13}{425}[/tex]

                                                                                        =  [tex]0.03058[/tex]

Hence The probability that the first two cards chosen are spades and the third card is red if the cards are chosen with replacement is 0.03125 . The probability that the first two cards chosen are spades and the third card is red if the cards are chosen without replacement is 0.03058

Final answer:

When the cards are chosen with replacement, the probability is 169/2704. When the cards are chosen without replacement, the probability is 39/884.

Explanation:

When the cards are chosen with replacement, the probability that the first two cards chosen are spades and the third card is red can be found by multiplying the probabilities of each event happening. There are 13 spades in a deck of 52 cards, so the probability of choosing a spade on the first draw is 13/52. Since the cards are chosen with replacement, the probability of choosing a spade on the second draw is also 13/52. The probability of choosing a red card on the third draw is 26/52, since there are 26 red cards in the deck. Therefore, the overall probability is (13/52) * (13/52) * (26/52) = 169/2704.

When the cards are chosen without replacement, the probability changes. The probability of choosing a spade on the first draw is 13/52. Since the first card is not replaced, there are now 51 cards left and 12 spades remaining. So the probability of choosing a spade on the second draw is 12/51. Finally, there are 25 red cards left out of 50 total cards, so the probability of choosing a red card on the third draw is 25/50. Therefore, the overall probability is (13/52) * (12/51) * (25/50) = 39/884.

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A golf ball is selected at random from a golf bag. If the golf bag contains 99 type A​ balls, 88 type B​ balls, and 44 type C​ balls, find the probability that the golf ball is not a type A ball.

Answers

Answer:

4/7

Step-by-step explanation:

To find the probability, we must first find the total number of golf balls. So 99+88+44 = 231 is our total. Now we must find the number of golf balls that are not a type A ball. This is just the total number of type B balls and type C balls. So 88+44 = 132. So the probability is 132/231. This simplifies to 12/21 or 4/7.

Find the intersection and union of the sets A and B given below:

A = { a, m, w, u, 7, y } , B = { w, y, g, 7}

Answers

Answer:

Intersection: {w, 7, y}

Union: {a, m, w, u, 7, y, g}

Step-by-step explanation:

The intersection of two or more sets are the elements they have in common, that means the elements that are in all the sets at the same time.

For example, "a" is in A but not in B, that's why is not in the intersection. On the other hand, "w" is in both sets, so it's in the intersection.

The union are all the elements that are in one set or the other, but we don't add the element twice if it's in both sets.  

For example, "a" is in A so we add it to the union. "w" is in A so we add it to the union but it's also in B, we don't add it again because it is already in the union.  

10. Sarah is planning to fence in her backyard garden. One side of the garden is 34 feet long, another side is 30 feet long, and the third side is 67 feet long.Find the perimeter of Sarah’s garden to determine the amount of fencing material needed.

A.262 ft.
B.68,340 ft.
C.250 ft.
D.131 ft.

Answers

Answer:

131ft is the amount of fencing material needed

Step-by-step explanation:

Perimeter is the distance around a shape: we have to sum all the distances

P = d1 + d2 + d3

P = 34 ft + 30 ft + 67 ft = 131 ft

Answer:

D. 131 ft.

Step-by-step explanation:

If Sarah is planning to fence in her backyard garden and one side of the garden is 34 feet long, another side is 30 feet long, and the third side is 67 feet long. The perimeter of Sarah’s garden to determine the amount of fencing material needed is 131 feet.

Using either the critical value rule or the p-value rule, if a one-sided null hypothesis is rejected at a given significance level, then the corresponding two-sided null hypothesis (i.e., the same sample size, the same standard deviation, and the same mean) will ______________ be rejected at the same significance level.

Answers

Final answer:

Rejecting a one-sided null hypothesis at a given significance level does not necessarily mean that the corresponding two-sided null hypothesis will also be rejected at the same significance level because one-sided tests and two-sided tests have different rejection regions.

Explanation:

Using either the critical value rule or the p-value rule, if a one-sided null hypothesis is rejected at a given significance level, then the corresponding two-sided null hypothesis (i.e., the same sample size, the same standard deviation, and the same mean) will not necessarily be rejected at the same significance level.

One-sided tests and two-sided tests have different rejection regions. For a one-sided test, the rejection region is all on one side of the sampling distribution, while for a two-sided test, the rejection regions are on both sides. If the test statistic falls in the rejection region for a one-sided test, it does not necessarily mean it will fall in the rejection region for the two-sided test, even at the same significance level.

Thus, even if you reject a one-sided null hypothesis at a given significance level, you cannot automatically reject the two-sided null hypothesis at the same level. You need to perform the appropriate statistical test.

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The mean per capita income is 24,787 dollars per annum with a variance of 169,744.What is the probability that the sample mean would differ from the true mean by greater than 42 dollars if a sample of 412 persons is randomly selected? Round your answer to four decimal places.

Answers

Final answer:

The question asks for the probability that the sample mean of income differs from the true mean by over 42 dollars for a sample size of 412. We use the Central Limit Theorem to approach this problem, calculating the standard error and Z-score to refer to the standard normal distribution to find the associated probability.

Explanation:

This question pertains to the field of statistics, more specifically, the Central Limit Theorem. The Central Limit Theorem says that if we take many samples from a population, the distribution of the sample means will approximate a normal distribution, regardless of the shape of the population distribution. We can use this theorem to calculate a probability related to the sample mean.

In this case, the population mean (μ) is a per capita income of 24,787 dollars and the population variance (σ²) is 169,744 dollars. We're asked to find the probability that the sample mean would differ from the true mean by more than 42 dollars if a sample of 412 persons is randomly selected.

The standard error of the sample mean is calculated by σ / sqrt(n), where σ is the standard deviation (sqrt(σ²)), and n is the sample size (412). After finding the standard error, we will calculate the Z-score of 42, which is the number of standard errors 42 is away from the mean. Calculating the Z-score is achieved by z = (X - μ) / SE, where X is the value of 42.

We can refer the calculated Z-score to the standard normal distribution to find the associated probability. However, since we are looking for the probability of a difference greater than 42, we want the probabilities in the tails of the distribution beyond our calculated Z-score, so it would be 1 minus this value.

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Every day a student randomly chooses a sandwich for lunch from a pile of wrapped sandwiches. If there are six kinds of sandwiches, how many different ways are there for the student to choose sandwiches for the seven days of a week if the order in which the sandwiches are chosen matters?

Answers

Final answer:

If the order matters and there are six kinds of sandwiches a student can choose for each of the seven days, there are 6^7, or 279,936, combinations possible. The calculation is based on the permutation rule of counting principle.

Explanation:

The student can select sandwiches in different ways following the rules of counting principle or more specifically permutation. Since the student can choose from six sandwiches each day, and this choice is made seven times (for seven days), the choice each day is an independent event because the choice of sandwich one day does not affect what he or she can choose the subsequent day.

 

The total number of ways the student can select sandwiches is given by raising the total number of choices (6) by the total number of days (7). So, there are 6^7 or 279,936 possible combinations of sandwiches for the week.

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Suppose you send out your newest "tweet" to your 5000 Twitter followers. You suspect that the change in the number of followers that have seen your tweet is proportional to the ratio of the number of followers that have seen the tweet and the number of followers that have not seen the tweet. If 10 followers have seen the tweet 5. after 1 minute, write a differential equation that models the number of followers that have seen the tweet, including any initial condition. [Do not solve the differential equation.]

Answers

Answer: Suppose we send out our newest "tweet" to our 5000 Twitter followers.

If 10 followers have seen the tweet after 1 minute, then the differential equation can be written as ;

Let us first assume that at time "t" , "n" followers have seen this tweet.

So, no. of follower who have not seen this tweet are given as : 5000 - n

ratio = [tex]\frac{n}{5000 - n}[/tex]

∴ we get ,

[tex]\frac{\delta x}{\delta t}[/tex] ∝  [tex]\frac{n}{5000 - n}[/tex]

[tex]\frac{\delta x}{\delta t}[/tex] = k×[tex]\frac{n}{5000 - n}[/tex]               ------ (1)

where k is the proportionality constant

At t = 0 , one follower has seen the tweet.

So n(0) = 0                                                                                   ------ (2)

So n(1) = 10                                                                                   ------ (3)

∴ equation (1), (2) and (3) together model the no. of followers that have seen the tweet.

The resistance,R(in ohms), Of a wire varies directly with the length, L(in cm) of the wire, and inversely with the cross-sectional area, A (in cm2). A 500 cm piece of wire with a radius of 0.2 cm has a resistance of 0.025 ohm. Find an equation that relates these variables.

Answers

Answer:

R = 2π×10⁻⁶ L / A

Step-by-step explanation:

Resistance varies directly with length and inversely with area, so:

R = kL/A

A round wire has cross-sectional area of:

A = πr²

Substituting:

R = kL/(πr²)

Given that R = 0.025 Ω when L = 500 cm and r = 0.2 cm:

0.025 = k (500) / (π (0.2²))

k = 2π×10⁻⁶

Therefore:

R = 2π×10⁻⁶ L / A

Answer:

0.00002 in the first blank. in the second blank its A

Step-by-step explanation:

A complex number, represented by z = x + iy, may also be visualized as a 2 by 2 matrix

(x y
-y x)

(a) Verify that addition and multiplication of complex numbers de ned via matrix opera-
tions are consistent with the usual addition and multiplication rules.
(b) What is the matrix representation corresponding to (x + iy)^-1?

Answers

Answer:

Step-by-step explanation:

A) Suppose that we have the complex numbers

[tex]z= x + iy \quad \text{and} \quad \\\\ \tilde{z}=\tilde{x} + i \tilde{y}[/tex]

Remember that to sum complex numbers, we sum the real parts of the two numbers to get the real part and the imaginary parts of the two numbers to get the imaginary part. Hence,  

[tex]z+\tilde{z} = (x + i y) + (\tilde{x} + i \tilde{y}) = (x + \tilde{x})+i (y+\tilde{y})[/tex]

On the other hand, if we sum the matrix visualizations of [tex]z \quad \text{and} \quad \tilde{z}[/tex] we get

[tex]\left[\begin{array}{cc}x &y\\-y&x\end{array}\right] + \left[\begin{array}{cc}\tilde{x}&\tilde{y}\\ -\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x + \tilde{x}& y + \tilde{y}\\-(y+\tilde{y})&x+\tilde{x}\end{array}\right][/tex]

which is the matrix visualization of [tex]z + \tilde{z}[/tex].

To multiply two complex numbers, we use the distributive law to multiplly and then separete the real part from the imaginary part

[tex]z \cdot \tilde{z}= (x + iy) \cdot (\tilde{x} + i \tilde{y})=(x \tilde{x} + i x \tilde{y} + i \tilde{x} y - y\tilde{y} ) = (x\tilde{x}-y\yilde{y})+i(x\tilde{y}+\tilde{x}y)[/tex]

Again, if we multiply the matrix visualizations of [tex]z \quad \text{and} \quad \tilde{z}[/tex] we get

[tex]\left[\begin{array}{cc}x&y\\-y&x\end{array}\right]\left[\begin{array}{cc}\tilde{x}&\tilde{y}\\-\tilde{y}&\tilde{x}\end{array}\right] = \left[\begin{array}{cc}x\tilde{x}-y\tilde{y}&x\tilde{y}+y\tilde{x}\\-y\tilde{x}-x\tilde{y}&x\tilde{x}-y\tilde{y}\end{array}\right][/tex]

which is the matrix viasualization of [tex]z\cdot\tilde{z}.[/tex]

B)  Since the usual matrix operations are consisten with the usual addition and multiplication rules in the complex numbers, we can use them to find the multiplicative inverses of a complex number [tex]z=x+iy[/tex].

We are looking for the complex number [tex]z^{-1}=(x+iy)^{-1}[/tex] which in terms of matrices is equivalent to find the matrix

[tex]\left[\begin{array}{cc}x&y\\-y & x\end{array}\right]^{-1}= \dfrac{1}{x^{2}+y^{2}} \left[\begin{array}{ccc}x&-y\\y&x\end{array}\right][/tex]    

Hence,

[tex]z^{-1}=\dfrac{1}{x^2 +y^2} (x-iy)=\dfrac{1}{|z|^2}(x-iy)[/tex]

Given that f(x)= x(squared)-3x Evaluate f(2+h)-f(2)/h

Answers

Answer:  The required value of the given expression is (h +1).

Step-by-step explanation:  We are given a function f(x) as follows :

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

We are to evaluate the following :

[tex]E=\dfrac{f(2+h)-f(2)}{h}.[/tex]

Substituting x = 2 + h in equation (i), we get

[tex]f(2+h)\\\\=(2+h)^2-3(2+h)\\\\=(4+4h+h^2)-6-3h\\\\=4+4h+h^2-6-3h\\\\=h^2+h-2[/tex]

and substituting x = 2 in equation (i), we get

[tex]f(2)=2^2-3\times2=4-6=-2.[/tex]

Therefore, the value of expression E can be evaluated as follows :

[tex]E\\\\\\=\dfrac{f(2+h)-f(2)}{h}\\\\\\=\dfrac{(h^2+h-2)-(-2)}{h}\\\\\\=\dfrac{h^2+h-2+2}{h}\\\\\\=\dfrac{h^2+h}{h}\\\\\\=\dfrac{h(h+1)}{h}\\\\=h+1.[/tex]

Thus, the required value of the given expression is (h +1).

Susan wants to make 2 square flags to
sell at a crafts fair. The fabric she wants
to buy is 6 meters wide. She doesn't
want any fabric left over. What's the
least amount of fabric she should buy?

Answers

She could buy 3 meters of fabric and cut that in half to have two 3 by 3 square flags or she could buy 12 meters and cut it in half to have two 6 by 6 square flags

Half of the students in an International School are Americans. One third of the remaining students are Europeans, and the rest are Australians. If there are 150 Australians, how many students are in the school?

Answers

Answer:

450 students

Step-by-step explanation:

Let s = number of students

1/2 are Americans  1/2s are Americans

s -1/2s = number of students left

1/2 s = number of students left

1/3 of the students left are Europeans

1/3 (1/2s) = 1/6s  are Europeans

The rest are Australians = 150

Students = Americans + Europeans + Australians

   s          =   1/2s          +    1/6s    + 150

Combine like terms

s    = 1/2*3/3 *s + 1/6s +150

s = 3/6s + 1/6s +150

s = 4/6s + 150

Subtract 4/6s from each side

s -4/6s = 150

6/6s - 4/6s = 150

2/6s = 150

1/3s = 150

Multiply each side by 3

3* 1/3s = 150*3

s = 450

There are 450 students

Answer:

There are 450 students.

Step-by-step explanation:

Let's call

s = total number of students

am = American students

eu = European students

au = Australian students

Half of the students in an International School are Americans. Then,

1/2 s = am [1]

One third of the remaining students are Europeans.

The remaining students are s - 1/2 s = 1/2 s

1/3 × (1/2 s) = 1/6 s = eu  [2]

There are 150 Australians.

au = 150 [3]

The total number of students is the sum of Americans, Europeans and Australians.

am + eu + au = s

Replacing [1], [2] and [3].

1/2 s + 1/6 s + 150 = s

2/3 s + 150 = s

1/3 s = 150

s = 450

3. Determine a lower bound for the radius of convergence of series solutions about each given point ro for the given different ial equation (a) y"4y6ry = 0; (b) (z2- 2r -3)"+ry'4y 0; ro 0, o4 ro4, ro= -4, ro = 0.

Answers

Answer:

Step-by-step explanation:

37

Although not so common now, consider a phone plan that does not have unlimited texting. Your base plan costs you $30 a month and each text sent or received is only 12 cents. Write an equation to describe the total monthly cost of your bill with x text messages and y cost.
In one month 217,000 messages were sent between two brothers in Philadelphia. What was their approximate bill? (This is a true story!)

Answers

Answer:

The bill will be $26070.

Step-by-step explanation:

Your base plan costs you $30 a month

And each text sent or received is only 12 cents or $0.12.

Let the messages sent be = x

So, cost for messages become = [tex]0.12x[/tex]

Let total bill cost be = y

Now, total bill will comprise of monthly plan charges plus message charges.

So, equation forms:

[tex]y=30+0.12x[/tex]

Now, given is that In one month 217,000 messages were sent between two brothers in Philadelphia.

So, their approximate bill will be :

[tex]y=30+0.12(217000)[/tex]

=>[tex]y=30+26040[/tex]

=> y = 26070

Therefore, bill will be $26070.

Final answer:

The answer explains how to write an equation for the total monthly cost of a phone plan based on text message usage and calculates the bill for 217,000 messages sent between two brothers in Philadelphia.

Explanation:

To write an equation for the total monthly cost of a phone plan with a base cost and a cost per text message:

Let x be the number of text messages sent or received.Let y be the total monthly cost.Using the given information, the equation is: y = $30 + 0.12x

For the specific case of 217,000 messages sent between two brothers:

Substitute x = 217,000 into the equation: y = $30 + 0.12(217,000)Calculate to find their approximate bill.

Check My Work (No more tries available) Solve the following word problem, rounding dollars to the nearest cent. The Flour Power Bakery makes 280 cherry cheesecakes at a cost of $2.51 each, If a spoilage rate of 30% is anticipated, at what price should the cakes be sold to achieye a 65% markup based on cost? Do not enter units in your answer 10.24 per cheesecake

Answers

Answer:

5.92 per cheesecake.

Step-by-step explanation:

The initial number cherry cheesecakes = 280,

Also, the cost of each cheesecakes = $ 2.51,

So, the total cost price = 280 × 2.51 = $ 702.8,

Markup = 65 %,

Thus, the total selling cost = 702.8 + 65% of 702.8 = $ 1159.62,

Now, the spoilage rate = 30 %,

So, the total new number of cheesecakes = 280 - 30% of 280 = 280 - 84 = 196,

Hence, the selling cost of each cheesecake = [tex]\frac{1159.62}{196}[/tex]

[tex]=\$ 5.91642857143[/tex]

[tex]\approx \$ 5.92[/tex]

Question 5: What is the radius of the circle with general form x2+y2-4y=21?
Question 5 options:

√21

4.58

5

25

Answers

Answer:

r=5 units

Step-by-step explanation:

we have

[tex]x^{2}+y^{2} -4y=21[/tex]

Convert the equation of the circle in standard form

Complete the square twice. Remember to balance the equation by adding the same constants to each side

[tex]x^{2}+(y^{2} -4y+4)=21+4[/tex]

[tex]x^{2}+(y^{2} -4y+4)=25[/tex]

Rewrite as perfect squares

[tex]x^{2}+(y-2)^{2}=25[/tex]

center (0,2)

radius r=5 units

Ann wants to mix a tomato sauce that is 17% sugar with a sauce that is 30% sugar to obtain 2.6 quarts of a tomato sauce that is 24% sugar. How much of each should she mix?

Answers

Answer:

she should mix 1.2 quarts of sauce with 17% sugar with 1.4 quarts of sauce with 30% sugar

Step-by-step explanation:

X= quarts of sauce # 1 ( 17% of sugar)

Y= quarts of sauce # 2( 30 % of sugar )

General Balance

[tex]X+Y=2.6[/tex]      EQ (1)

[tex]Y=2.6-X\\[/tex]

SUGAR BALANCE

[tex]0.17X+0.3Y=2.6*0.24\\[/tex]   EQ (2)

replace Y from eq(1) into eq(2)

[tex]0.17X+0.3(2.6-X)=0.624\\0.17X+0.78-0.3X=0.624\\-0.13X=0.624-0.780\\X=\frac{-0.156}{-0.13} \\X= 1.2\\Y=2.6-1.2=1.4[/tex]

If a company employed 50 people in 1995, and tripled their employment by 2005, how many total people would be employed if there are 40% more employees than in 2005 by 2015?

Answers

Answer:

210

Step-by-step explanation:

In 2005 the employee was tripple of 1995 which is 50×3=150. So number of employees in 2005=150.

In 2015 number of employees were 40% of employees in 2005.

40% of 150 =150×40÷100 = 60,

Therefore number of employees in 2015 =150+60= 210 employees.

Answer:

210 employees in 2015

Step-by-step explanation:

In order to solve this we first have to calculate the number of employees that there were in 2005, since the problem says that they tripled their employees from 1995, that meas three times 50, meaning 150 employees in 2005, if by 2015 the number of employees had increase in 40% we just do a simple rule of three, 150 being the 100% and trying to calculate the 140%:

[tex]\frac{150}{100}= \frac{x}{15} \\x=\frac{150*15}{100}\\ x= 210[/tex]

So now we know that there were 210 employees in 2015.

What does the symbol R indicate in an experimental design? a. observation b. experimental variable C. comparison groups d. random assignment

Answers

Answer:

R indicates Random Assignment

Step-by-step explanation:

In designing an experiment, R is the symbol used for assigning the subjects randomly to different groups through randomization. It is called random assignment and It is the option (d) random assignment. It is achieved by using chance procedure ( random number generator, by flipping coin, throwing ). This ensures that all the subjects have equal chance of placing in all the groups.

Option (a) Symbol O is used for Observations. (So option (a) is wrong answer)

Option (b) Symbols x , y are used as Observational variables (So option (b) is wrong answer)

Option (c) Comparison groups is wrong answer as R is not used for that.

Final answer:

The symbol R in experimental design represents random assignment, a technique used to ensure that groups in an experiment are initially equivalent and thereby allow researchers to make causal inferences.

Explanation:

In experimental design, the symbol R indicates random assignment. Random assignment is a process used to create initial equivalence between the groups in an experiment, which is crucial for allowing researchers to draw causal conclusions. In this process, participants are assigned to different groups or conditions on a random basis, such as by drawing numbers or using a random number generator.

Random assignment minimizes the effects of lurking variables, ensuring that before the experimental manipulation occurs, the groups can be considered equivalent on various potential confounding factors. When individuals are selected into groups denoted by R, we can generally classify these designs as experimental, as opposed to nonexperimental designs where nonrandom assignment (NR) is indicated.

For example, in a study where the size of tableware is manipulated to see its effect on food consumption, the psychologist would use random assignment to ensure that both treatment groups are comparable prior to the intervention, which is the manipulation of the independent variable.

1.prove that the following three functions are linearly dependent

f1=x^2; f2(x)=1-x^2; f3=2+x^2

Answers

Proof:

Given any functions [tex]f_{1}(x),f_{2}(x),f_{3}(x)[/tex] they are linearly dependent if we can find values of [tex]c_{1},c_{2},c_{3}[/tex] such that

[tex]c_{1}f_{1}(x)+c_2f_{2}(x)+c_{3}f_{3}(x)=0[/tex]

Using the given functions in the above equation we get

[tex]c_{1}f_{1}(x)+c_2f_{2}(x)+c_{3}f_{3}(x)=0\\\\c_{1}x^{2}+c_{2}(1-x^{2})+c_{3}(2+x^{2})=0\\\\\Rightarrow (c_{1}-c_{2}+c_{3})x^{2}+c_{1}+c_{2}+2c_{3}=0[/tex]

This will be satisfied if and only if

[tex]c_1-c_2+c_3=0,c_1+c_2+2c_3=0[/tex]

Solving the equations we get

[tex]c_1+2c_3=-c_2\\\\c_1+c_1+2c_3+c_3=0\\2c_1+3c_3=0[/tex]

Since we have 3 variables and 2 equations thus we will get many solutions  

one being if we put [tex]c_3=1[/tex] we get

[tex]c_1+2c_3=-c_2\\\\c_1+c_1+2c_3+c_3=0\\2c_1+3c_3=0\\\\c_1=\frac{-3}{2}\\\\c_2=\frac{-1}{2}[/tex]

Thus we have [tex]c_1=\frac{-3}{2},c_2=\frac{-1}{2},c_3=1[/tex] as one solution. Hence the given functions are linearly dependent.

Assume the random variable x is normally distributed with mean muμequals=8787 and standard deviation sigmaσequals=55. Find the indicated probability. ​P(x less than<7979​)

Answers

Answer:  0.4404

Step-by-step explanation:

Let the random variable x is normally distributed .

Given : Mean : [tex]\mu=\ 87[/tex]

Standard deviation : [tex]\sigma= 55[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = 79

[tex]z=\dfrac{79-87}{55}\approx-0.15[/tex]

The p-value = [tex]P(x<79)=P(z<-0.15)[/tex]

[tex]=0.4403823\approx0.4404[/tex]

Hence, the required probability : [tex]P(x<79)=0.4404[/tex]

A multiple choice test contains 10 questions, each with 3 possible answers (of which only one is correct). If a student answers each question by rolling a die and choosing the first answer if the die shows 1 or 2, the second answer if the die shows 3 or 4, or the third answer if the die shows 5 or 6, what is the probability that the student will get exactly 6 correct answers? more than 6 correct answers?

Answers

Answer:

1. The probability that the student will get exactly 6 correct answers is [tex]\frac{1120}{19683}[/tex].

2. The probability that the student will get more than 6 correct answers is [tex]\frac{43}{2187}[/tex].

Step-by-step explanation:

From the given information it is clear that

The total number of equations (n) = 10

The probability of selecting the correct answer (p)= [tex]\frac{1}{3}[/tex]

The probability of selecting the incorrect answer (q)= [tex]1-p=1-\frac{1}{3}=\frac{2}{3}[/tex]

According to the binomial distribution, the probability of selecting r items from n items is

[tex]P=^nC_rp^rq^{n-r}[/tex]

where, p is probability of success and q is the probability of failure.

The probability that the student will get exactly 6 correct answers is

[tex]P(r=6)=^{10}C_6(\frac{1}{3})^6(\frac{2}{3})^{10-6}[/tex]

[tex]P(r=6)=210(\frac{1}{3})^6(\frac{2}{3})^{4}=\frac{1120}{19683}[/tex]

Therefore the probability that the student will get exactly 6 correct answers is [tex]\frac{1120}{19683}[/tex].

The probability that the student will get more than 6 correct answers is

[tex]P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{10-7}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{10-8}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{10-9}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{10-10}[/tex]

[tex]P(r>6)=^{10}C_7(\frac{1}{3})^7(\frac{2}{3})^{3}+^{10}C_8(\frac{1}{3})^8(\frac{2}{3})^{2}+^{10}C_9(\frac{1}{3})^9(\frac{2}{3})^{1}+^{10}C_{10}(\frac{1}{3})^{10}(\frac{2}{3})^{0}[/tex]

[tex]P(r>6)=120\times \frac{8}{59049}+45\times \frac{4}{59049}+10\times \frac{2}{59049}+1\times \frac{1}{59049}=\frac{43}{2187}[/tex]

Therefore the probability that the student will get more than 6 correct answers is [tex]\frac{43}{2187}[/tex].

ONE HALF OF AN ANNUALLY SALARY OF $35,700. IS APPROXIMATELY

Answers

Answer: Approximately $18,000

Step-by-step explanation:

Since you are approximating, the 7 indicates the need to round the original number up to 36,000. Then divide by 2 since you want half, thus getting $18,000.

Huck and Jim are waiting for a raft. The number of rafts floating by over intervals of time is a Poisson process with a rate of λ = 0.4 rafts per day. They agree in advance to let the first raft go and take the second one that comes along. What is the probability that they will have to wait more than a week? Hint: If they have to wait more than a week, what does that say about the number of rafts in a period of 7 days?

Answers

Answer: 0.0081

Step-by-step explanation:

Let X be the number of rafts.

Given : The mean number of rafts floating : [tex]\lambda=0.4[/tex] rafts per day .

Then , for 7 days the number of rafts = [tex]\lambda_1=\lambda\times 7=0.4\times7=2.8[/tex] rafts per day .

The formula to calculate the Poisson distribution is given by :_

[tex]P(X=x)=\dfrac{e^{-\lambda_1}\lambda_1^x}{x!}[/tex]

Now, the  probability that they will have to wait more than a week is given by :-

[tex]P(X>7)=1-P(X\leq7)\\\\=1-(P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(5)+P(6)+P(7))\\\\=1-(\dfrac{e^{-2.8}2.8^{0}}{0!}+\dfrac{e^{-2.8}2.8^{1}}{1!}+\dfrac{e^{-2.8}2.8^{2}}{2!}+\dfrac{e^{-2.8}2.8^{3}}{3!}+\dfrac{e^{-2.8}2.8^{4}}{4!}+\dfrac{e^{-2.8}2.8^{5}}{5!}+\dfrac{e^{-2.8}2.8^{6}}{6!}+\dfrac{e^{-2.8}2.8^{7}}{7!})\\\\=1-0.991869258012=0.008130741988\approx0.0081[/tex]

Hence, the required probability : 0.0081

solve l 3x-7 l = 2
and solve l 2x-5 l< or equal to 8
the lines (l) signifies absolute values therefore l 3x-7 l is the absolute value of that and same thing for 2x-5

Answers

Answer:

1) Solutions are x = 3 and x = 5/3

2) Solution are x ≤ 13/2 and  x ≤ -3/2

Step-by-step explanation:

1) Given absolute inequality,

|3x-7| = 2

⇒ 3x - 7 = ± 2

⇒ 3x = 7 ± 2

[tex]\implies x=\frac{7\pm 2}{3}[/tex]

[tex]x=\frac{7+2}{3}\text{ or }x=\frac{7-2}{3}[/tex]

[tex]\implies x = 3\text{ or }x=\frac{5}{3}[/tex]

2) l 2x-5 l ≤ 8

⇒  2x-5  ≤ ±8

⇒   2x ≤ 5 ± 8

[tex]\implies x\leq \frac{5\pm 8}{2}[/tex]

[tex]x\leq \frac{5+8}{2}\text{ or }x\leq \frac{5-8}{2}[/tex]

[tex]\implies x \leq \frac{13}{2}\text{ or }x\leq-\frac{3}{2}[/tex]

Answer:

Answer:

1) Solutions are x = 3 and x = 5/3

2) Solution are x ≤ 13/2 and  x ≤ -3/2

Step-by-step explanation:

1) Given absolute inequality,

|3x-7| = 2

⇒ 3x - 7 = ± 2

⇒ 3x = 7 ± 2

2) l 2x-5 l ≤ 8

⇒  2x-5  ≤ ±8

⇒   2x ≤ 5 ± 8

Suppose babies born in a large hospital have a mean weight of 4095 grams, and a standard deviation of 569 grams. If 130 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by greater than 42 grams? Round your answer to four decimal places.

Answers

Answer:

[tex]P =0.3998[/tex]

Step-by-step explanation:

Let [tex]{\displaystyle {\overline{x}}}[/tex] be the average of the sample, and the population mean will be [tex]\mu[/tex]

We know that:

[tex]\mu = 4095[/tex] gr

Let [tex]\sigma[/tex] be the standard deviation and n the sample size, then we know that the standard error of the sample is:

[tex]E=\frac{\sigma}{\sqrt{n}}[/tex]

Where

[tex]\sigma=569[/tex]

[tex]n=130[/tex]

In this case we are looking for:

[tex]P(|{\displaystyle{\overline{x}}}- \mu|>42)[/tex]

This is:

[tex]{\displaystyle{\overline{x}}}- \mu>42[/tex] or [tex]{\displaystyle{\overline{x}}}- \mu<-42[/tex]

[tex]P=P({\displaystyle{\overline{x}}}- \mu>42)+ P({\displaystyle{\overline{x}}}- \mu<-42)[/tex]

Now we get the z score

[tex]Z=\frac{{\displaystyle{\overline{x}}}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

[tex]P=P(z>\frac{42}{\frac{569}{\sqrt{130}}}) + P(z<-\frac{42}{\frac{569}{\sqrt{130}}})[/tex]

[tex]P=P(z>0.8416) + P(z<-0.8416)[/tex]

Looking at the tables for the standard nominal distribution we get

[tex]P =0.1999+0.1999[/tex]

[tex]P =0.3998[/tex]

Two friends visited Washington D.C. for the weekend. Person 1 rode the subway one stop, 5 times at the peak fare price and four times at the off peak fare price for a total cost of $17.40. Person 2 rode with Person 1, 2 times at the peak fare price and three times at the off peak fair price for a total cost of $9.20. How much was the peak fare price?

Answers

Answer:

The peak fare price is $2,2

Step-by-step explanation:

Considering

peak fare price= xoff peak fare price = y

Person 1  

5x + 4y= 17,40

Person 2

2x + 3y= 9,20

Isolating the x

2x= 9,20-3y

x=(9,20-3y)/2=4,6 -1,5y

x=4,6 -1,5y

Replacing in the other equation

5 (4,6 -1,5y) + 4y=17,40

23-7,5y+4y=17,40

23-17,40=7,5y-4y

5,6=3,5y

y=1,60 (this is the off peak fare)

x= 4,60-1,5y

x= 4,60-1,5*1,60

x=2,2 (this is the peak fare)

Final answer:

To determine the peak fare price, we formed a system of linear equations based on the given data. By using the elimination method, we solved the system to find that the peak fare price is $2.20.

Explanation:

We have a system of linear equations representing the total costs of peak and off-peak fares for two friends using public transportation in Washington D.C. Let's denote the peak fare as P and the off-peak fare as Op. According to the problem, we are given the following two equations based on the trips taken and the total cost paid by each person:

5P + 4Op = $17.40 (Person 1)

2P + 3Op = $9.20 (Person 2)

To find the peak fare price, we need to solve this system of equations. We can use a variety of methods, including substitution or the elimination method.

Step 1 - Use the Elimination Method:

Multiply the second equation by 2 to align the Op terms:

4P + 6Op = $18.40 (after multiplying by 2)

Step 2 - Subtract the resulting equation from the first person's total cost:

5P + 4Op - (4P + 6Op) = $17.40 - $18.40

P - 2Op = -$1.00

Step 3 - Solve for P:

Isolate P by adding 2Op to both sides:

P = 2Op - $1.00

Now, substitute P into the second equation from Person 2's total cost:

2(2Op - $1.00) + 3Op = $9.20

4Op - $2.00 + 3Op = $9.20

7Op - $2.00 = $9.20

Add $2.00 to both sides to get 7Op alone:

7Op = $11.20

Divide by 7 to find the off-peak fare:

Op = $11.20 / 7

Op = $1.60

Use the value of Op to find P:

P = 2($1.60) - $1.0

P = $3.20 - $1.00

P = $2.20

The peak fare price is $2.20.

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