Find the missing lengths and angle measures in kite ABCD

Answer:

The two numbers are 37 and 26.

Step-by-step explanation:

This question can be solved by a simple system of equations.

Building the system:

I am going to say that our numbers are x and y.

Two numbers total 63

This means that [tex]x + y = 63[/tex]

difference of 11

This means that [tex]x - y = 11[/tex].

Solving the system

[tex]x + y = 63[/tex]

[tex]x - y = 11[/tex]

Using the addition method

x + x + y - y = 63 + 11

2x = 74

x = 37

[tex]x - y = 11[/tex]

[tex]y = x - 11[/tex]

[tex]y = 37 - 11 = 26[/tex]

The two numbers are 37 and 26.

Answer:

The total 2 numbers are 37 and 26

hope i helped

Answer:

a.  Inferential statistics

b. Descriptive statistics

Step-by-step explanation:

statistics can be categorized into descriptive an inferential statistics. descriptive statistics makes use of a set of data for numerical calculations  and provides conclusion based on those numerical calculation. this data could be collected using tables,graphs, and other means of data representation.

Inferential statistics however come up with conclusions and assumptions base on a sample data.

Examples of descriptive statistics are mean, median, mode, quartile, percentile. Thus option B is descriptive.

Option A however is inferential statistics since some assumptions were made based on the effect of garlic on blood pressure.

Answer:

There is a 54.88% probability that there are no surface flaws in an auto's interior.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

[tex]e = 2.71828[/tex] is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.06 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel.

For one square foot, we have 0.06 flaws.

So for 10 square feet, the mean is [tex]\mu = 10*0.06 = 0.6[/tex]

(a) What is the probability that there are no surface flaws in an auto's interior

This is P(X = 0).

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-0.6}*(0.6)^{0}}{(0.6)!} = 0.5488[/tex]

There is a 54.88% probability that there are no surface flaws in an auto's interior.

Answer:

The information a measure of variability for the variable competitiveness conveys is

The first option is correct

a.) Do all teenage girls have the same amount of competitiveness?

The second option is correct

b.) How spread out are the values for competitiveness among teenage girls?

Step-by-step explanation:

The first option is correct

a.) Do all teenage girls have the same amount of competitiveness?

The second option is correct

b.) How spread out are the values for competitiveness among teenage girls?

The third option is NOT correct. The given problem is a measure of dispersion not central tendency.

c.) What is the central tendency for the variable competitiveness among teenage girls?

The fourth option is also NOT correct. The problem given does not talk about a specific value of the competitiveness variable but rather it talks about the variability of the competitiveness variable

d.) What is the most common value for the variable competitiveness among teenage girls?

Answer:

1) 0.3

2) 0.64

3) 0.75

4) 0.33                              

Step-by-step explanation:

We are given the following probabilities. We have to probability of that event not happening.

That is we have to find the complement of the event,

Complement of event:

1. P(E) = 0.7

[tex]P(E') = 1 - P(E) = 1 - 0.7 =0.3[/tex]

2. P(E) = 0.36

[tex]P(E') = 1 - P(E) = 1 - 0.36 = 0.64[/tex]

3. P(E) = 1/4

[tex]P(E') = 1 = P(E) = 1 -\dfrac{1}{4} = \dfrac{3}{4} = 0.75[/tex]

4. P(E) = 2/3

[tex]P(E') = 1 = P(E) = 1 -\dfrac{2}{3} = \dfrac{1}{3} = 0.33[/tex]

Answer:

0.1673 (16.73%)

Step-by-step explanation:

The probability that a fair coin is chosen is 2/3 . if the fair coin is chosen the probability of getting 3 heads is determined by binomial distribution:

P(3 heads in 10 flips )=B(n=10,p=0.5,X=3) =0.1171

But if a coin that is not fair is chosen , then the probability of getting 3 heads also follows a binomial distribution but with p=0.3

P(3 heads)=B(n=10,p=0.3,X=3)=0.2668

Finally the probability is

P= 2/3*0.1171 + 1/3* 0.2678 = 0.1673 (16.73%)

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by [tex]d(t)=\frac{t^{2}}{6}+4t[/tex] where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

[tex]V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}[/tex]

so let's take the first interval:

(a) [1.95, 2.05]

[tex]V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}[/tex]

we get that:

[tex]d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125[/tex]

[tex]d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167[/tex]

so:

[tex]V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s[/tex]

(b) [1.995, 2.005]

[tex]V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}[/tex]

we get that:

[tex]d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831[/tex]

[tex]d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835[/tex]

so:

[tex]V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s[/tex]

(c) [1.9995, 2.0005]

[tex]V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}[/tex]

we get that:

[tex]d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358[/tex]

[tex]d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358[/tex]

so:

[tex]V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s[/tex]

(d) [2, 2.00001]

[tex]V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}[/tex]

we get that:

[tex]d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333[/tex]

[tex]d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333[/tex]

so:

[tex]V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s[/tex]

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

The average velocity is computed for each interval using the average velocity formula and the given position function, d(t) = 3t^2 + 4t + 6. Instantaneous velocity is calculated by differentiating the position function and substituting the given time point.

To find the average velocity, we need to use the formula: average velocity = (final position - initial position) / (final time - initial time). For the position function given, d(t) = 3t^2 + 4t + 6, substitute the time values given in the intervals to find the position at those times, then use those values to calculate the average velocity.

For example, for the first interval [1.95, 2.05], d(1.95) = 3*(1.95)^2 + 4*1.95 + 6 and d(2.05) = 3*(2.05)^2 + 4*2.05 + 6. Subtract these position values and the time values then divide the result.

The instantaneous velocity is calculated by taking the slope of the tangent line at the specified point on the position vs time curve, which is found using the derivative of the position function. The derivative of d(t) = 3t^2 + 4t + 6 is d'(t) = 6t + 4. So, at t = 2, the instantaneous velocity would be d'(2) = 6*2 + 4.

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Answer:

a) Figure attached

b) [tex] P(X<130 min)= F(130) = \frac{130-120}{20}=0.5[/tex]

c) [tex] P(X>135) =1-P(X<135) = 1- F(135) = 1- \frac{135-120}{20}= 0.25[/tex]

d) [tex] E(X) =\frac{a+b}{2}= \frac{120+140}{2}= 130min[/tex]

Step-by-step explanation:

2 hr = 120 min

2hr 20 min = 120+20 = 140 min

Let X the random variable that represent "flight time". And we know that the distribution of X is given by:

[tex]X\sim Uniform(120 ,140)[/tex]

For this case the density function for A is given by:

[tex] f(X) = \frac{1}{b-a}= \frac{1}{20} , 120\leq X \leq 140 [/tex]

And [tex] f(X)= 0[/tex] for other case

Part a

For this case we can see the figure attached. We see that the probability density function is defined between 120 and 140 minutes.

Part b

The arrival time is 2hr 5 min = 125 min. So then 5 minutes leate means 130 min

For this case we want to find this probability:

[tex] P(X<130 min)[/tex]

And we can use the cumulative distribution function for X given by:

[tex] F(X) = \frac{x-120}{140-120} = \frac{x-120}{20} , 120\leq X \leq 140 [/tex]

And if we use this formula we got:

[tex] P(X<130 min)= F(130) = \frac{130-120}{20}=0.5[/tex]

Part c

For this case more than 10 minutes late means 125 +10 = 135 min or more, so we want this probability:

[tex] P(X>135)[/tex]

And using the complement rule we have:

[tex] P(X>135) =1-P(X<135) = 1- F(135) = 1- \frac{135-120}{20}= 0.25[/tex]

Part d

The expected value for the uniform distribution is given by:

[tex] E(X) =\frac{a+b}{2}= \frac{120+140}{2}= 130min[/tex]

Answer:

(a) 0.0001 or 0.01%

(b) 0.01 or 1%

Step-by-step explanation:

Since there are 10 possible numeric digits (from 0 to 9), and there is only one correct digit, there is a 1 in 10 change of getting each digit right.

The probability that if you forget your PIN, then you can guess the correct sequence

(a) at random:

[tex]P=(\frac{1}{10})^4=0.0001=0.01\%[/tex]

(b) when you recall the first two digits.

[tex]P=(\frac{1}{10})^2=0.01=1\%[/tex]

The probability of guessing a four-digit PIN code correctly is 1/10,000 if guessed randomly and 1/100 if the first two digits are recalled correctly.

To find the probability of guessing a four-digit PIN code correctly, we need to consider the number of possible combinations and the number of favorable outcomes. There are 10 possible choices for each digit (0-9), so there are 10^4 = 10,000 possible four-digit PIN codes.

(a) If you forget your PIN and guess it randomly, there is only one correct sequence out of the 10,000 possibilities. Therefore, the probability of guessing the correct sequence is 1/10,000.

(b) If you recall the first two digits correctly, there are still 10 choices for each of the remaining two digits. So, the probability of guessing the correct sequence in this case is 1/10^2 = 1/100.

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The velocity of the ball after 3 seconds will be 29.43 meters per second.

The analysis of mathematical representations is algebra, and the handling of those symbols is logic.

Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground.

Then the velocity of the ball after 3 seconds will be given as,

We know that the first equation of motion.

v = u - gt

Where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time in seconds. Then we have

v = 0 - 9.81 x 3

v = -29.43 m/s

The velocity of the ball after 3 seconds will be 29.43 meters per second.

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Final answer:

The velocity of the ball after 3 seconds can be found using Galileo's law of free fall. By calculating the average velocity over a brief time interval, we can approximate the instantenous velocity. The velocity of the ball after 3 seconds is therefore m/s.

Explanation:

The velocity of the ball after 3 seconds can be found using Galileo's law of free fall. According to the given information, the distance fallen after t seconds is given by the equation s(t) = 4.9t², where s(t) is the distance fallen in meters and t is the time in seconds. To find the velocity, we can approximate it by computing the average velocity over a brief time interval. We can calculate the average velocity from t = 3 to t = 3.1 by finding the change in position divided by the time elapsed. Plugging the values into the equation, we get:

Average velocity = (s(3.1) - s(3)) / 0.1 = (4.9(3.1)² - 4.9(3)²) / 0.1 = m/s

Therefore, the velocity of the ball after 3 seconds is m/s, rounded to one decimal place.

Answer:

a) Binomial probability distribution. Only two outcomes possible for each machine, with independent probabilities.

b) [tex]P(X = 4) = 0.00045[/tex]

c) [tex]P(X \geq 4) = 0.00046[/tex]

d) [tex]E(X) = 0.5[/tex]

e) [tex]V(X) = 0.45[/tex]

Step-by-step explanation:

For each machine, there is only two possibilities. On a given day, either they will break down, or they will not. The probabilities for each machine breaking down are independent. So we use the binomial probability distribution to solve this problem.

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.

The expected value of the binomial distribution is:

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

The variance of the binomial distribution is:

[tex]V(X) = np(1-p)[/tex]

In this problem we have that:

[tex]n = 5, p = 0.1[/tex]

a. What is the appropriate probability distribution for x? Explain how x satisfies the properties of the distribution.

Binomial probability distribution. Only two outcomes possible for each machine, with independent probabilities.

b. Compute the probability that 4 machines will break down.

This is [tex]P(X = 4)[/tex].

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

[tex]P(X = 4) = C_{5,4}.(0.1)^{4}.(0.9)^{1} = 0.00045[/tex]

c. Compute the probability that at least 4 machines will break down.

This is

[tex]P(X \geq 4) = P(X = 4) + P(X = 5)[/tex]

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

[tex]P(X = 4) = C_{5,4}.(0.1)^{4}.(0.9)^{1} = 0.00045[/tex]

[tex]P(X = 5) = C_{5,5}.(0.1)^{5}.(0.9)^{0} = 0.00001[/tex]

[tex]P(X \geq 4) = P(X = 4) + P(X = 5) = 0.00045 + 0.00001 = 0.00046[/tex]

d. What is expected number of machines that will break down in a day?

[tex]E(X) = np = 5*0.1 = 0.5[/tex]

e. What is the variance of the number of machines that will break down in a day?

[tex]V(X) = np(1-p) = 5*0.1*0.9 = 0.45[/tex]

The random variable x follows a binomial distribution with n=5 and p=0.1. The probability of exactly 4 machines breaking down is approximately 0.00045, while the probability of at least 4 machines breaking down is approximately 0.00046. The expected number of machines that will break down is 0.5 and the variance is 0.45.

The appropriate probability distribution for x is the binomial distribution. A binomial distribution has two outcomes (a machine breaks down, or it doesn't), a fixed number of trials (5 machines), and a constant probability of success (0.1, a machine breaks down). x satisfies all these properties.

To compute the probability that 4 machines will break down, we can use the binomial theorem: P(x=k) = C(n,k) * (p^k) * (1-p)^(n-k). Here, n=5, k=4, p=0.1. The calculation gives us approximately 0.00045 as the probability that exactly 4 machines will break down.

To compute the probability that at least 4 machines will break down, we calculate the sum of the probabilities that 4 and 5 machines will break down. It's approximately 0.00046.

The expected number of machines that will break down in a day is calculated by n*p, which gives us 0.5 machines.

The variance of the number of machines that will break down in a day is np(1-p), which gives us 0.45.

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Answer:

a) 0.06

b) 0.3

c) 0.44                                        

Step-by-step explanation:

We are given the following in the question:

A and B are two independent events.

P(A) = 0.3

P(B) = 0.2

a) P(A intersect ​B)

[tex]P(A\cap B) = P(A)\times P(B) = 0.3\times 0.2 = 0.06[/tex]  

b) P(A|B)

[tex]P(A|B) = \dfrac{P(A\cap B)}{P(B)} = \dfrac{0.06}{0.2} = 0.3[/tex]

c) P(A union ​B)

[tex]P(A\cup B) = P(A) + P(B) - P(A\cap B)\\P(A\cup B) = 0.3 + 0.2 - 0.06 = 0.44[/tex]

Answer: 160.4040

Step-by-step explanation:

Here , the claim is "The standard deviation of pulse rates of adult males is more than 11 bpm." , i.e. [tex]\sigma>11[/tex]

We use Chi-square test statistic for the test statistic to test the standard deviations :

[tex]\chi^2=\dfrac{(n-1)s^2}{\sigma^2}[/tex] , where s= sample standard deviation , [tex]\sigma[/tex] = population standard deviations and n = sample size.

As per given , n=153 , s= 11.3

Then, Required test statistic will be :

[tex]\chi^2=\dfrac{(153-1)(11.3)^2}{11^2}=\dfrac{152\times127.69}{121}\approx160.4040[/tex]

Hence, the value of the test statistic. =  160.4040

Answer:

(a) {RR,BB,RB,RY,BY}

(b) {R1 R2, R1 B1, R1 B2, R1 Y, R2 R1, R2 B1, R2 B2, R2 Y, B1 R1, B1 R2, B1 B2, B1 Y, B2 R1, B2 B1, B2 R2, B2 Y, Y R1, Y R2, Y B1, Y B2}.

Step-by-step explanation:

This question can be answered in two ways :

(a) When order of picking marbles does not matter.

(b) When order of picking marbles does matter.

Red marbles in the bag = 2

Blue marbles in the bag = 2

Yellow marbles in the bag = 1

Now since we have to select two marbles from the bag, the sample cases will be:

In short Sample space = {RR,BB,RB,RY,BY} where R = Red , B = Blue and Y = Yellow.

    2. Now I will explain what will be the sample space for the experiment  when order of picking marbles does matter.

For this, First give numbers to the balls in bag i.e.,

First Red ball in the bag = R1

Second Red ball in the bag = R2

First Blue ball in the bag = B1

Second Blue ball in the bag = B2

Yellow ball in the bag = Y

Now the cases for sample spaces when two marbles are selected will be :

{R1 R2, R1 B1, R1 B2, R1 Y, R2 R1, R2 B1, R2 B2, R2 Y, B1 R1, B1 R2, B1 B2, B1 Y, B2 R1, B2 B1, B2 R2, B2 Y, Y R1, Y R2, Y B1, Y B2}.

Answer:

1300 mm  

Step-by-step explanation:

You want to convert metres to millimetres, so you multiply the metres by a conversion factor:

1.3 m × conversion factor = x mm

1 m = 1000 mm.

So,  the conversion factor is either (1 m/1000 mm) or (1000 mm/1 m).

We choose the latter, because it has the desired units on top.

The calculation becomes

[tex]\text{Length} = \text{1.3 m} \times \dfrac{\text{1000 mm}}{\text{1 m}} = \textbf{1300 mm}[/tex]

Final answer:

a. The probability that your bid will be accepted is 60%. b. The probability that your bid will be accepted is 10%. c. To maximize the probability of getting the property, you should bid $14,599. d. To maximize expected profit, bid $14,599 with an expected profit of $700.50.

Explanation:

a. To calculate the probability that your bid will be accepted, we need to find the probability that your bid is higher than the competitor's bid. Since the competitor's bid, x, is uniformly distributed between $10,400 and $14,600, the probability that the competitor's bid is less than $12,000 is given by:

P(x < 12,000) = (12,000 - 10,400) / (14,600 - 10,400) = 0.4

Therefore, the probability that your bid will be accepted is 1 - P(x < 12,000) = 1 - 0.4 = 0.6, or 60%.

b. Using the same method, the probability that your bid will be accepted when you bid $14,000 is:

P(x < 14,000) = (14,000 - 10,400) / (14,600 - 10,400) = 0.9

Therefore, the probability that your bid will be accepted is 1 - P(x < 14,000) = 1 - 0.9 = 0.1, or 10%.

c. To maximize the probability that you get the property, you should bid an amount that is slightly higher than the competitor's bid, but still below $14,600. This is because if your bid is equal to the competitor's bid, there is a 50% chance that the seller will choose your bid and a 50% chance that the seller will choose the competitor's bid. Therefore, to maximize your chances, you should bid $14,599.

d. If your objective is to maximize the expected profit, you should consider the probability of winning the property and the profit you will make if you win. Since you know someone is willing to pay you $16,000 for the property, the expected profit can be calculated as:

Expected Profit = Probability of Winning × (Selling Price - Bidding Amount)

According to part (c), if you bid $14,599, the probability of winning is 0.5. Therefore, the expected profit is:

Expected Profit = 0.5 × (16,000 - 14,599) = $700.50

a. Probability of bid acceptance when bidding $12,000 is approximately 0.38.

b. Probability of bid acceptance when bidding $14,000 is approximately 0.86.

c. To maximize chances, bid $12,500.

d. For maximum expected profit, bid $13,200, with an expected profit of approximately $3,000.

Let's break down the problem step by step:

a. Probability of bid acceptance when bidding $12,000:

Given that the competitor's bid ( x ) is uniformly distributed between $10,400 and $14,600, we need to find the probability that our bid is higher than $10,400 but less than $14,600.

[tex]\[ P(10400 < x < 12000) = \frac{12000 - 10400}{14600 - 10400} = \frac{1600}{4200} \][/tex]

b. Probability of bid acceptance when bidding $14,000:

[tex]\[ P(10400 < x < 14000) = \frac{14000 - 10400}{14600 - 10400} = \frac{3600}{4200} \][/tex]

c. To maximize the probability of winning, we need to find the midpoint of the range $10,400 to $14,600, which is $12,500. So, we should bid $12,500.

d. Expected profit when bidding $12,500:

If someone is willing to pay $16,000 for the property, and we win the bid, our profit would be $16,000 - $12,500 = $3,500. The probability of winning the bid when bidding $12,500 is the same as calculated in part c. So, the expected profit would be:

[tex]\[ E(Profit) = (Probability \ of \ winning) \times (Profit \ if \ won) \][/tex]

[tex]\[ E(Profit) = \frac{3600}{4200} \times 3500 \][/tex]

Now, let's calculate:

a. [tex]\[ P(10400 < x < 12000) = \frac{1600}{4200} \approx 0.38 \][/tex]

b.[tex]\[ P(10400 < x < 14000) = \frac{3600}{4200} \approx 0.86 \][/tex]

c. Midpoint: $12,500

d. [tex]\[ E(Profit) = \frac{3600}{4200} \times 3500 \approx 3000 \][/tex]

So, to maximize expected profit, the bid should be $13,200.

Answer:

1 spider, 3 cows, 5 birds.

Step-by-step explanation:

Cows (C): 4 legs, 2 eyes.

Birds (B): 2 legs, 2 eyes.

Spiders (S): 8 legs, 4 eyes.

The number of legs and eyes are given, respectively by:

[tex]30 = 4C+2B+8S\\20 = 2C+2B+4S[/tex]

Multiplying the second equation by -2 and adding it to the first one gives us the number of birds:

[tex]30 -40= 4C-4C+2B-4B+8S-8S\\-10 = -2B\\B=5[/tex]

Rewriting the original equations with B =5:

[tex]30 = 4C+2*5+8S\\20 = 2C+2*5+4S\\20 = 4C+8S\\10 = 2C+4S\\C=5-2S[/tex]

Since the number of cows cannot be negative, and both C and S must be odd numbers, the only possible value of S is 1:

[tex]C=5-2S\\S=1\\C=3[/tex]

There are 3 cows, 5 birds, and 1 spider in the field.

To solve this problem, we need to establish equations based on the given information

1. Define Variables

Let c be the number of cows,

b be the number of birds,

and s be the number of spiders

2. Create Equations

The total number of eyes is given to be 20, and knowing that spiders have 4 eyes, birds have 2 eyes, and cows have 2 eyes, we get:

2c + 2b + 4s = 20

Simplifying this, we get:

c + b + 2s = 10  (i)

The total number of legs is given to be 30, with cows having 4 legs, birds having 2 legs, and spiders having 8 legs, we get:

4c + 2b + 8s = 30

Simplifying this, we get:

2c + b + 4s = 15  (ii)

3. Solve Simultaneous Equations

We can subtract equation (i) from (ii):

(2c + b + 4s) - (c + b + 2s) = 15 - 10

c + 2s = 5  (iii)

Substituting (iii) into (i):

(5 - 2s) + b + 2s = 10

5 + b = 10

b = 5;

We know from (iii) that:

c + 2s = 5

Let’s check for odd values for s:

If s = 1, then c = 5 - 2(1) = 3 (valid since cows = 3)

Hence, the solution is:

c = 3, b = 5, s = 1

Checking:

Total eyes = 2(3) + 2(5) + 4(1) = 6 + 10 + 4 = 20 (correct)

Total legs = 4(3) + 2(5) + 8(1) = 12 + 10 + 8 = 30 (correct)

There are 3 cows, 5 birds, and 1 spider in the field.

Answer:

1) 0.9772

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 40000, \sigma = 5000[/tex]

What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

This is 1 subtracted by the pvalue of Z when X = 30000. So

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

[tex]Z = \frac{30000 - 40000}{5000}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228.

1 - 0.0228 = 0.9772

0.9772 that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000

Final answer:

Using the normal distribution, the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000 is approximately 0.9772.

Explanation:

The probability of a randomly selected individual with an MBA degree getting a starting salary of at least $30,000 is determined using the normal distribution with a mean of $40,000 and a standard deviation of $5,000. You calculate the z-score for $30,000, which is (30,000 - 40,000) / 5,000 = -2.

Then, you look this z-score up in the standard normal distribution table (or use a calculator equipped with normal distribution functions) to find the probability of getting a value greater than -2. This probability corresponds to the area to the right of the z-score, which is essentially 1 minus the cumulative probability up to the z-score.

According to standard normal distribution tables, the cumulative probability of a z-score of -2 is approximately 0.0228. Therefore, the probability of getting at least $30,000 is

1 - 0.0228, which equals approximately 0.9772.

Comparing this to available choices, option 1) 0.9772 is the correct answer.

Answer:

705,432 ways

Step-by-step explanation:

Since no girl will be left out once both teams are selected when selecting the varsity team, the junior varsity team is automatically composed by the players not selected, the number of ways to select both teams is:

[tex]n = \frac{22!}{(22-11)!11!} \\n=705,432[/tex]

There are 705,432 ways to divide the girls into a varsity team and a junior varsity team.

Answer:

d. 14:11

Step-by-step explanation:

There is a 7 in 10 chance that Mike does his homework today and a 8 in 10 chance he will do it tomorrow,  the probability that he will do it both today and tomorrow is:

[tex]P = \frac{7}{10}* \frac{8}{10}\\P=\frac{56}{100}=0.56[/tex]

The odds are the probability that an event occurs over the probability that it does not occur:

[tex]Odds = \frac{0.56}{1-0.56}=\frac{56}{44}=\frac{14}{11}[/tex]

The odds that he will do it both today and tomorrow are 14:11.

The odds that he will do it both today and tomorrow is 14:11.

Probability that he would do the assignement today and tomorrow = probability he would do the assignment today x probability he would do the assignment tomorrow

0.7 x 0.8 = 0.56

Odds he would do the assignment today and tomorrow = 0.56/ (1 - 0.56) = 0.56 / 0.44

= 14 : 11

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Answer: 3 to the 4 power

Step-by-step explanation:

Answer:

3^4

because it is 3 times it self 4 times

Answer:

300

Step-by-step explanation:

20% of 1,500 = 300

(Source: Google)

Answer:

300

Step-by-step explanation:

Answer:

account after 2 years of simple interest at 10%: (2,500 * 0.10 * 2) + 2500 = 3,000  

Note that simple interest only pays interest on the original balance, NOT on the accrued (paid) interest....

a) 5 more years at 10% simple: (2500 * 0.10 * 5) + 3,000 = $4,250

or

5 years compound interest on $3k: 3,000(1.07^5) = $4,207.66

b) TOTAL of 10 years

simple interest: (2500 * 0.10 * 10) + 2500 = 5,000

compound interest: only 8 years remain of the total 10 year time horizon

3000(1.07^8) = $5,154.56

Step-by-step explanation:

After the initial 2 years with simple interest, moving the money to the account with compound interest yields a higher amount whether you liquidate in the next 5 years or keep it for a total of 10 more years.

First, let's calculate the amount you will have after 2 years with a simple interest of 10%. Using the formula for simple interest, I = PRT, where P is the principal amount ($2,500), R is the rate of interest (10% as a decimal, 0.10), and T is the time in years (2): I = $2,500 * 0.10 * 2 = $500. Therefore, your total amount after 2 years would be $2,500 + $500 = $3,000.

To decide whether you should move your money, we need to calculate the final amount after the next 5 years and 10 years using compound interest formula, A = P(1 + r/n)^(nt), where P is the principal amount ($3,000), r is the annual interest rate (7% as a decimal, 0.07), n is the number of times interest applied per time period (1, for annual compounding), and t is the time the money is invested for.

(a) For 5 more years: A = $3,000 * (1 + 0.07/1)^(1*5) = $4,209.24.
(b) For a total of 10 more years: A = $3,000 * (1 + 0.07/1)^(1*10) = $5,922.58.

Therefore, If you intend to liquidate in five more years or you're confident your money will stay on deposit for a total of ten more years after the initial 2 years of simple interest, moving the money to the compound interest account is a good move as it results in a higher amount in both scenarios.

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Answer:

a) If we construct a stemplot for the data we have this:

Stem     Leaf

1       |     9

2      |     2 3 9

3      |     0 0 1 3 5

4      |     2

Notation: 1|9 means 19

b) [tex]29.4-2.26\frac{6.75}{\sqrt{10}}=24.58[/tex]    

[tex]29.4-2.26\frac{6.75}{\sqrt{10}}=34.22[/tex]    

So on this case the 95% confidence interval would be given by (24.58;34.22)    

Step-by-step explanation:

For this case we have the following data:

30 30 42 35 22 33 31 29 19 23

Part a

If we construct a stemplot for the data we have this:

Stem     Leaf

1       |     9

2      |     2 3 9

3      |     0 0 1 3 5

4      |     2

Notation: 1|9 means 19

As we can see the distribution is a little assymetrical to the right since we have not to much values on the right tail. But we can approximate roughly the distribution symmetric and with no outliers.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

The confidence interval for the mean is given by the following formula:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=29.4[/tex]

The sample deviation calculated [tex]s=6.75[/tex]

In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:

[tex]df=n-1=10-1=9[/tex]

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.26[/tex]

Now we have everything in order to replace into formula (1):

[tex]29.4-2.26\frac{6.75}{\sqrt{10}}=24.58[/tex]    

[tex]29.4-2.26\frac{6.75}{\sqrt{10}}=34.22[/tex]    

So on this case the 95% confidence interval would be given by (24.58;34.22)    

Answer:

This is called Simpson's Paradox.

Therefore the correct option is A.) Simpson's Paradox.

Step-by-step explanation:

i) This is called Simpson's Paradox.

ii) If there are trends that tend to appear in several different groups of data which apparently either disappear or tend to reverse when these data groups are combined.

This is an example of Simpson's Paradox, where a trend appears in different groups of data but disappears or reverses when the groups are combined.

This is an example of Simpson's Paradox. Simpson's Paradox occurs when a trend appears in different groups of data but disappears or reverses when the groups are combined. In this case, although students who studied more slept less in both groups (those who went out and those who stayed in the dorm), when the groups are combined, those who studied more also slept more.

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Answer:

Mr. Rosenbloom 1 bottle of gel will last for 20 days.

Step-by-step explanation:

Given:

Amount of gel used every morning = 500 mL

Amount of gel in 1 bottle = 10 liters.

We need to find the number of days 1 bottle of gel will last for.

Solution:

Now we know that;

1 liter =1000 mL

10 liter = 10000 mL

Now we now that;

500 mL of gel is used = 1 day

10000 mL of gel will be used = Number of days 10000 mL of gel will last.

By Using Unitary method we get;

Number of days 10000 mL of gel will last = [tex]\frac{10000}{500}=20\ days[/tex]

Hence Mr. Rosenbloom 1 bottle of gel will last for 20 days.

Answer:

y = 2x

Step-by-step explanation:

Claire is hungry. She buys 2 donuts each costing x $. How much should she pay?

Since one donut costs x $ 2 donuts cost 2x $.

Therefore, total amount Claire should pay, call it y = 2x.

Hence, we have y = 2x.

Final answer:

A real-life word problem for the equation y=2x could be renting a bike at a rate of $2 per hour, in which the cost (y) is proportional to the rental time (x). This scenario shows a linear relationship with the total cost increasing directly with the rental time.

Explanation:

A real-life word problem for the equation y=2x could involve a situation where the cost y (in dollars) to hire a bike is proportional to the number of hours x you rent it for. If it costs $2 per hour to rent a bike, then the total cost is represented by the equation y=2x. For example, renting the bike for 3 hours would cost y=2(3)=6 dollars.

Applying this equation, it follows that as the time of rental increases, the cost increases at a constant rate. This represents a linear relationship between time and cost.

Example Calculation:

Let x represent the time in hours you rent a bike.

The total cost y is twice the rental time (since it's $2 per hour).

To find the cost of renting a bike for 5 hours, substitute x=5 into the equation: y=2(5).

The total cost would be y=$10.

The equation y=2x is used here to model a simple proportional relationship where one variable increases directly as the other does. It does not represent a quadratic, hyperbolic, or any other non-linear relationship.

Answer:

y=\frac{ -\frac{\sqrt[9]{18} Г (10/9, -\frac{19t^{18}}{18})}{19^{10/9} {e^{\frac{19t^{18}}{18}}}

Step-by-step explanation:

From exercise we have:

C=0

y'+19t^{17} y=t^{19}

We know that a linear differential equation is written in the standard form:

y' + a(t)y = f(t)

we get that: a(t)=19t^{17} and f(t)=t^{19}.

We know that the integrating factor is defined by the formula:

u(t)=e^{∫ a(t) dt}

⇒ u(t)=e^{∫ 19t^{17} dt} = e^{\frac{19t^{18}}{18}}

The general solution of the differential equation is in the form:

y=\frac{ ∫ u(t) f(t) dt +C}{u(t)}

y=\frac{ ∫ e^{\frac{19t^{18}}{18}} · t^{19} dt + 0}{e^{\frac{19t^{18}}{18}}}

y=\frac{ -\frac{\sqrt[9]{18} Г (10/9, -\frac{19t^{18}}{18})}{19^{10/9} {e^{\frac{19t^{18}}{18}}}

Answer:

a. 68% of the workers will earn between $47300 and $69700.

b. 2.5% of workers will earn above $89000

c. Approximately 0

Step-by-step explanation:

The standard normal distribution curve in the attached graph is used to solve this question.

a. The value $47300 is a standard deviation below the mean i.e. 58500-11200=47300. While $69700 is a standard deviation above the mean. I.e. 58500+12000=69700.

Between the first deviation below and above the mean, you have 34%+34%=68% of the salary earners within this range. So we have 68%of staffs earning within this range

b. The second standard deviation above the mean is $80900. i.e. 58500+11200+11200=$80900

We have 50%+13.5%+2.5%= 97.5% earning below $80900. Therefore, 100-97.5= 2.5% of the workers earn above this amount.

c. From the Standard Deviation Rule, the probability is only about (1 -0 .997) / 2 = 0.0015 that a normal value would be more than 3 standard deviations away from its mean in one direction or the other. The probability is only 0.0002 that a normal variable would be more than 3.5 standard deviations above its mean. Any more standard deviations than that, and we generally say the probability is approximately zero.

To answer the question, we use z-scores and a z-table to find the percentages of workers who earn within certain ranges or above certain amounts.

To answer this question, we can use the concept of the standard normal distribution. First, we convert the given earnings into z-scores by subtracting the mean and dividing by the standard deviation. With these z-scores, we can then use a z-table to find the percentage of workers who earn within a certain range or above a certain amount.

a. To find the percentage of workers who earn between $47,300 and $69,700, we need to convert these values into z-scores and find the area between these two z-scores on the z-table.

b. To find the percentage of workers who earn more than $80,900, we need to convert this value into a z-score and find the area to the right of this z-score on the z-table.

c. To determine how likely it is that someone earns more than $100,000, we need to convert this value into a z-score and find the area to the right of this z-score on the z-table.

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Answer: for replacement total games = 47   more than 3.5 hours   = 8 required proportion = 8/47 = 0.170212 for regular.

Step-by-step explanation:

1. What proportion of the 47 games officiated by replacement referees lasted for at least 3.5 hours (210 minutes)? What proportion of the 42 games officiated by regular referees lasted for this long?  

2. What proportion of the 47 games officiated by replacement referees lasted for less than 3 hours (180 minutes)? What proportion of the 42 games officiated by regular referees lasted for this long?  

3. Would you say that either type of referee tended to have longer games than the other on average, and if so, which type of referee tended to have longer games and by about how much on average?

i. Regular Referees have longer game times with games about 185 minutes on average.

ii. The game lengths for both referees seem to be the same.

iii. Replacement Referees have longer game times with games about 195 minutes on average.

iv. It is too hard to tell from the graph which type of referee has longer games.

4.  Would you say that either type of referee tended to have more variability in game durations? If so, which type of referee tended to have more variability?

i. The two types of referees had the same amount of variability in the game lengths.

ii. There is no way to tell which group of referees had more variability in the game lengths.

iii. The regular referees tended to have more variability in the game lengths.

iv. The replacement referees tended to have more variability in the game lengths.

For replacement total games = 47   more than 3.5 hours   = 8 required proportion = 8/47 = 0.170212 for regular.