Find the sample space for the experiment.
You select two marbles (without replacement) from a bag containing two red marbles, two blue marbles, and one yellow marble. You record the color of each marble.

Answer:

P-value or Probability value is the exact percentage where test statistics lie.The criterion for rejecting the null hypothesis using the​ P-value approach is that if P-value < Level of significance , then we reject our null hypothesis

Step-by-step explanation:

P-value or Probability value is the exact percentage where test statistics lie.

It also tells the probability of obtaining extreme results corresponding to our level of significance keeping in state that our null hypothesis is true or correct.

The criterion for rejecting the null hypothesis using the​ P-value approach is that if P-value < Level of significance , then we reject our null hypothesis i.e.  

Suppose P-value is 2.33% and Level of significance is 5%, then we will reject our null hypothesis as 2.33% < 5%.

On the other hand, if P-value > Level of significance , then we cannot reject or accept our null hypothesis.

A p-value is a measure of the strength of evidence against the null hypothesis in a statistical hypothesis test. The criterion for rejecting the null hypothesis using the p-value approach is to compare the calculated p-value to a predetermined significance level.

A p-value is a measure of the strength of evidence against the null hypothesis in a statistical hypothesis test. It represents the probability of observing a test statistic as extreme or more extreme than the one observed, assuming that the null hypothesis is true. In other words, it quantifies how likely it is for the observed data to have occurred by chance if the null hypothesis is true.

The criterion for rejecting the null hypothesis using the p-value approach is to compare the calculated p-value to a predetermined significance level (usually denoted as alpha). If the p-value is smaller than alpha, we reject the null hypothesis, indicating that there is enough evidence to support the alternative hypothesis. Alternatively, if the p-value is greater than or equal to alpha, we fail to reject the null hypothesis, suggesting that there is not enough evidence to support the alternative hypothesis.

To simplify 25 to the power of 1/2 using the radical form, we equate this to the square root of 25. Since 5 squared is 25, the square root of 25 is 5, hence 25^1/2 equals 5.

To simplify 25 to the power of 1/2 using the radical form, we look at the properties of exponents and radicals. Exponentiation and radicals are inverse operations, so the expression 251/2 is equivalent to the square root of 25 because squaring a number and then taking the square root of it returns the original number.

So, 251/2 = √25. Since the square of 5 is 25, the square root of 25 is 5. Therefore, 251/2 = 5.

Answer:

(a) 0.06681

(b) 0.86638

(c)  [tex]\sigma[/tex] = 0.000214

Step-by-step explanation:

We are given that the diameter of the dot produced by a printer is normally distributed with a mean diameter of 0.002 inch and a standard deviation of 0.0004 inch i.e.;   [tex]\mu[/tex] = 0.002 inch        and        [tex]\sigma[/tex] = 0.0004

Also,   Z = [tex]\frac{X -\mu}{\sigma}[/tex] ~ N(0,1)

(a) Let X = diameter of a dot

    P(X > 0.0026 inch) = P( [tex]\frac{X -\mu}{\sigma}[/tex] > [tex]\frac{0.0026 -0.002}{0.0004}[/tex] ) = P(Z > 1.5) = 1 - P(Z <= 1.5)

                                                                          = 1 - 0.93319 = 0.06681

(b) P(0.0014 < X < 0.0026) = P(X < 0.0026) - P(X <= 0.0014)

    P(X < 0.0026) = P( [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{0.0026 -0.002}{0.0004}[/tex] ) = P(Z < 1.5) = 0.93319

    P(X <= 0.0014) = P( [tex]\frac{X -\mu}{\sigma}[/tex] <= [tex]\frac{0.0014 -0.002}{0.0004}[/tex] ) = P(Z <= -1.5) = 1 - P(Z <= 1.5)

                                                                     = 1 - 0.93319 = 0.06681

Therefore, P(0.0014 < X < 0.0026) = 0.93319 - 0.06681 = 0.86638 .

(c) P(0.0014 < X < 0.0026) = 0.995

    P( [tex]\frac{0.0014 -0.002}{\sigma}[/tex] < [tex]\frac{X -\mu}{\sigma}[/tex] < [tex]\frac{0.0026 -0.002}{\sigma}[/tex] ) = 0.995

    P( [tex]\frac{ -0.0006}{\sigma}[/tex] < Z < [tex]\frac{0.0006}{\sigma}[/tex] ) = 0.995

    P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - P(Z <= [tex]\frac{-0.0006}{\sigma}[/tex] ) = 0.995

    P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - (1 - P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) ) = 0.995

     2 * P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) - 1 = 0.995

           P(Z < [tex]\frac{0.0006}{\sigma}[/tex] ) = 0.9975

On seeing the z table we observe that at critical value of x = 2.81 we get the probability area of 0.9975 i.e.;

                  [tex]\frac{0.0006}{\sigma}[/tex] = 2.81      ⇒  [tex]\sigma[/tex] = 0.000214

Therefore, 0.000214 standard deviation of diameters is needed so that the probability in part (b) is 0.995 .

The probabilities of obtaining a given diameter is found from the z-table,

given that the dot produced by the printer are normally distributed.

Reasons:

The mean diameter, μ = 0.002

The standard deviation, σ = 0.0004

A. The probability that the diameter exceeds 0.0026 inch

Solution;

[tex]\displaystyle z-score,\ Z= \mathbf{\dfrac{x-\mu }{\sigma }}[/tex]

At x = 0.0026 inch, we have;

[tex]\displaystyle Z=\dfrac{0.0026-0.002 }{0.0004 } = 1.5[/tex]

P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668

The probability that the diameter of a dot exceeds 0.0026 inch = 0.0668

B. The probability that the diameter is less than 0.0026 inch = 0.9332

The z-score for a diameter of x = 0.0014 inch is given as follows;

[tex]\displaystyle Z=\mathbf{\dfrac{0.0014-0.002 }{0.0004 }} = -1.5[/tex]

P(Z < -1.5) = 0.0668

Therefore, the probability that the diameter is between 0.0014 and 0.0026 inch is given as follows;

P(0.0014 < x < 0.0026) = 0.9332 - 0.0668 = 0.8664

The probability that the diameter is between 0.0014 and 0.0026 = 0.8664

C. For the probability in part (b) to be 0.995, we have;

For a probability of 0.995, the z-score ≈ 2.575

[tex]\displaystyle P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right)- P\left(Z < \dfrac{0.0014-0.002 }{\sigma } \right)= 0.995[/tex]

Therefore;

[tex]\displaystyle \mathbf{ P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right)} = 0.995 + \frac{1 - 0.995}{2} = 0.9975[/tex]

From the z-table, we get;

[tex]\displaystyle P\left(Z < \dfrac{0.0026-0.002 }{\sigma } \right) = 0.9975[/tex]

The z-score with a probability of 0.9975 = 2.81

Which gives;

[tex]\displaystyle \left( \dfrac{0.0026-0.002 }{\sigma } \right) = 2.81[/tex]

[tex]\displaystyle \sigma = \left( \dfrac{0.0026-0.002 }{2.81} \right) = \mathbf{2.135 \times 10^{-4}}[/tex]

The standard deviation of the diameters needed so that the probability in part (b) is 0.995 is 2.135 × 10⁻⁴

Learn more z-score here:

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Answer:

Chi-square value = 2

Step-by-step explanation:

Given data:

Frequency:      Tall     Short

                         30       20

Null hypothesis ( as it is already given so there is no difference between observed and expected values)

Ratio:    1:1 (50% expected frequency of each tall and short pea-plant)

Solution:

Phenotype  Observed  Expected O-E  (O-E)²  (O-E)[tex]^{2/E}[/tex]

                          O                  E

Short                20               25        -5       25        1                          

TALL                  30               25         5        25       1

TOTAL                                                                       2

So, from all these calculations using expected and observed values we get chi-square value equal to 2.

Final answer:

To test the null hypothesis that the deviation from a 1:1 phenotypic ratio is due to chance in a cross between a heterozygous tall and a homozygous short pea plant, the Chi-square value is calculated to be 2.

Explanation:

The question relates to a cross between a heterozygous tall pea plant and a homozygous short pea plant, with the intention to calculate the Chi-square value to test the null hypothesis that the observed deviation from a 1:1 ratio is due to chance. In this scenario, the expectation is a 1:1 phenotypic ratio, meaning 25 tall and 25 short plants out of 50 offspring.

To calculate the Chi-square (χ²) value, the formula is χ² = Σ  ( (observed - expected)² / expected ), where Σ symbolizes the sum of calculations for each category. For tall plants, the calculation is ((30-25)² / 25) = (5² / 25) = 1 and for short plants, the calculation is ((20-25)² / 25) = (5² / 25) = 1. Therefore, the total χ² value is 1 + 1 = 2.

Answer:

Step-by-step explanation:

There are 200 dice out of which 199 are fair

Prob for 6 in one special die = 1 and

Prob for 6 in other die = 1/6

A1- drawing special die and A2 = drawing any other die

A1 and A2 are mutually exclusive and exhaustive

P(A1) = 1/200 and P(A2) = 199/200

B = getting 6

i) Required probability

= P(A1/B) = [tex]\frac{P(A1B)}{P(A1B)+P(A2B)} \\[/tex]

P(A1B) = [tex]\frac{1}{200} *1 = \frac{1}{200}[/tex]

P(A2B) = [tex]\frac{199}{200}*\frac{1}{6}=\frac{199}{1200}[/tex]

P(B) = [tex]\frac{205}{1200} =\frac{41}{240}[/tex]

P(A1/B) = [tex]\frac{1/200}{41/240} =\frac{7}{205}[/tex]

P(A2/B) = [tex]\frac{199/1200}{41/240} =\frac{199}{205}[/tex]

Answer:

(0.215,0.33)

Step-by-step explanation:

The 99% confidence interval can be calculated as

[tex]p- z_{\frac{\alpha }{2} } \sqrt{\frac{pq}{n} } <P<p+z_{\frac{\alpha }{2} } \sqrt{\frac{pq}{n} }[/tex]

Where p is the estimated sample proportion that can be calculated as

p=x/n

where x=109 and n=400

p=109/400=0.2725

q=1-p=1-0.273=0.7275

[tex]z_{\frac{\alpha }{2} } =z_{\frac{\0.01 }{2} }=z_{0.005}=2.5758[/tex]

The 99% confidence interval  is

[tex]0.2725-2.5758 \sqrt{\frac{0.2725(0.7275)}{400} } <P<0.2725+2.5758 \sqrt{\frac{0.2725(0.7275)}{400} }[/tex]

0.2725-2.5758(0.022262 )< P < 0.2725+2.5758(0.022262)

02725-0.057343 < P < 0.2725+0.057343

0.215157 < P < 0.329843

Rounding the obtained answer to three decimal places

0.215 < P < 0.33

Thus, the 99% confidence interval for the proportion of fatal accidents that involved the use of a cell phone is (0.215,0.33).

We are 99% confident that population the proportion of fatal accidents that involved the use of a cell phone will lie in this interval (0.215,0.33).

Answer:

Explanation:

The image attached has the mentioned histogram.

Such histogram presents the relative frequencies for the clases [0,1], [1,2],[2,3], [4,5], and [5,6] Silver in ppm.

Only the rectangle for the class [3,4] is missing.

The height of each rectangle is the relative frequency of the corresponding class.

The relative frequencies must add 1, because each relative frequency is calculated dividing the absolute class frequency by the total number in the sample; hence, the sum of all the relative frequencies is equal to the total absolute class frequencies divided by the same number, yielding 1.

In consequence, you can sum all the known relative frequencies and subtract from 1 to get the missing relative frequency, which is the height of the missing rectangle.

1. Sum of the known relative frequencies:

2. Missing frequency:

3. Conclusion:

Answer:

1) 0.375

2) 0.375

3) 0.5

4) 0.5

5) 0.875

6) 0.5                          

Step-by-step explanation:

We are given the following in the question:

Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

[tex]\text{Probability} = \displaystyle\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}[/tex]

1. The probability of getting exactly one tail

P(Exactly one tail)

Favorable outcomes ={HHT, HTH, THH}

[tex]\text{P(Exactly one tail)} = \dfrac{3}{8} = 0.375[/tex]

2. The probability of getting exactly two tails

P(Exactly two tail)

Favorable outcomes ={ HTT,THT, TTH}

[tex]\text{P(Exactly two tail)} = \dfrac{3}{8} = 0.375[/tex]

3. The probability of getting a head on the first toss

P(head on the first toss)

Favorable outcomes ={HHH, HHT, HTH, HTT}

[tex]\text{P(head on the first toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

4. The probability of getting a tail on the last toss

P(tail on the last toss)

Favorable outcomes ={HHT,HTT,THT,TTT}

[tex]\text{P(tail on the last toss)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

5. The probability of getting at least one head

P(at least one head)

Favorable outcomes ={HHH, HHT, HTH, HTT, THH, THT, TTH}

[tex]\text{P(at least one head)} = \dfrac{7}{8} = 0.875[/tex]

6. The probability of getting at least two heads

P(Exactly one tail)

Favorable outcomes ={HHH, HHT, HTH,THH}

[tex]\text{P(Exactly one tail)} = \dfrac{4}{8} = \dfrac{1}{2} = 0.5[/tex]

Answer:

the amount of water that is left in the tank after 10 min is 98 L

Step-by-step explanation:

since the water drains off at rate that is proportional to the water present

(-dV/dt) = k*V , where k= constant

(-dV/V) = k*dt

-∫dV/V) = k*∫dt

-ln V/V₀=k*t

or

V= V₀*e^(-k*t) , where V₀= initial volume

then since V₁=0.7*V₀ at t₁= 3 min

-ln V₁/V₀=k*t₁

then for t₂= 10 min we have

-ln V₂/V₀=k*t₂

dividing both equations

ln (V₂/V₀) / ln (V₁/V₀) =(t₂/t₁)

V₂/V₀ = (V₁/V₀)^(t₂/t₁)

V₂=V₀ *  (V₁/V₀)^(t₂/t₁)

replacing values

V₂=V₀ *  (V₁/V₀)^(t₂/t₁)  = 200 L * (0.7)^(10min/5min) = 98 L

then the amount of water that is left in the tank after 10 min is 98 L

The problem represents a case of exponential decay. Initially, 30% of water, or 60 liters, leaks out in 5 minutes, leaving 140 liters in the tank. Assuming the same rate of leakage, another 30% of water or 42 liters will leak out in the next 5 minutes, leaving 98 liters in the tank after 10 minutes.

In this problem, we are dealing with a situation involving exponential decay due to the water leakage which happens at a rate proportional to the amount of water present in the tank.

First, let's consider the 30% of water that leaks out in the first 5 minutes from a 200-liter tank. This amounts to 60 liters (200 * 0.30), which leaves 140 liters (200 - 60) of water in the tank after 5 minutes.

Now, since the decrease of water is proportional to the amount of water present, this implies an exponential decay over time. Given that the amount of water decreased by 30% in the first 5 minutes, it's reasonable to assume that it will decrease by the same percentage in the next 5 minutes as well.

So, the amount of water left in the tank after 10 minutes would be 98 liters (140 * 0.70).

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Answer:

Assuming population data

[tex] \sigma = \sqrt{0.000354}=0.0188[/tex]

Assuming sample data

[tex] s = \sqrt{0.000425}=0.0206[/tex]

Step-by-step explanation:

For this case we have the following data given:

736.352, 736.363, 736.375, 736.324, 736.358, and 736.383.

The first step in order to calculate the standard deviation is calculate the mean.

Assuming population data

[tex]\mu = \frac{\sum_{i=1}^6 X_i}{6}[/tex]

The value for the mean would be:

[tex]\mu = \frac{736.352+736.363+736.375+736.324+736.358+736.383}{6}=736.359[/tex]

And the population variance would be given by:

[tex] \sigma^2 = \frac{\sum_{i=1}^6 (x_i-\bar x)}{6}[/tex]

And we got [tex] \sigma^2 =0.000354[/tex]

And the deviation would be just the square root of the variance:

[tex] \sigma = \sqrt{0.000354}=0.0188[/tex]

Assuming sample data

[tex]\bar X = \frac{\sum_{i=1}^6 X_i}{6}[/tex]

The value for the mean would be:

[tex]\bar X = \frac{736.352+736.363+736.375+736.324+736.358+736.383}{6}=736.359[/tex]

And the population variance would be given by:

[tex] s^2 = \frac{\sum_{i=1}^6 (x_i-\bar x)}{6-1}[/tex]

And we got [tex] s^2 =0.000425[/tex]

And the deviation would be just the square root of the variance:

[tex] s = \sqrt{0.000425}=0.0206[/tex]

To determine the standard deviation of given length measurements, calculate the mean, find squared differences, average them, and take the square root. The standard deviation of the measurements is 0.01879 feet.

Determining the Standard Deviation of Measurements

To find the standard deviation of the given length measurements, follow these steps:

Calculate the mean of the measurements.

Find the squared differences between each measurement and the mean.

Compute the average of the squared differences.

Take the square root of that average.

Step-by-Step Calculation

→ List of measurements: 736.352, 736.363, 736.375, 736.324, 736.358, 736.383.

→ Calculate the mean:

= (736.352 + 736.363 + 736.375 + 736.324 + 736.358 + 736.383) / 6

= 736.3591667

Find the squared differences:

→ Calculate the average of the squared differences:

= (0.00005256 + 0.00001464 + 0.00024811 + 0.00123264 + 0.00000136 + 0.00057044) / 6

= 0.000353292

Take the square root to find the standard deviation:

= √(0.000353292)

= 0.01879 feet

Answer:

a) 0.80

b) 0.45

c) 0.55

Step-by-step explanation:

Given P(A) = 0.20 and P(B) = 0.35

Applying probability of success and failure; P(success) + P( failure) = 1

a) probability that the component does not fail the test = The component does not fail a particular test [P(success)] = 1 - P(A)

= 1 - 0.20 = 0.80

b) probability that the component works perfectly well

= P( the component works perfectly well) - P(component shows strain but does not fail test)

= 0.80 - 0.35 = 0.45

c) probability that the component either fails or shows strain in the test = 1 - P(the component works perfectly well)

= 1 - 0.45 = 0.55

This question is based on the concept of probability. Therefore, the answers are, (a) 0.80, (b) 0.45 and (c) 0.55.

Given:

Event A occurs with probability P(A) =  0.20, and event B occurs with probability  P(B) = 0.35.

According to the question,

Given P(A) = 0.20 and P(B) = 0.35,

As we know that,  probability of success and failure,

⇒ P(success) + P( failure) = 1

a) Probability that the component does not fail the test = The component does not fail a particular test

= P(success) = 1 - P(A)

= 1 - 0.20 = 0.80

b) Probability that the component works perfectly well

= P( the component works perfectly well) - P(component shows strain but does not fail test)

= 0.80 - 0.35 = 0.45

c) Probability that the component either fails or shows strain in the test = 1 - P(the component works perfectly well)

= 1 - 0.45 = 0.55

Therefore, the answers are, (a) 0.80, (b) 0.45 and (c) 0.55.

For more details, prefer this link:

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Answer:

a) 0.15

b) 0.2

c) 0.6

Step-by-step explanation:

We are given the following in the question:

A: Stopping at first signal

B: Stopping at second signal

P(A) = 0.35

P(B) = 0.55

Probability that he must stop at at least one of the two signals is 0.75

[tex]P(A\cup B) = 0.75[/tex]

a) P(at both signals)

[tex]P(A\cup B) = P(A) + P(B) - P(A\cap B)\\0.75 = 0.35 + 0.55 - P(A\cap B)\\P(A\cap B) = 0.35 + 0.55 - 0.75 = 0.15[/tex]

0.15 is the probability that motorist stops at both signals.

b) P(at the first signal but not at the second one)

[tex]P(A\cap B') = P(A) - P(A\cap B)\\P(A\cap B') = 0.35 - 0.15 = 0.2[/tex]

0.2 is the probability that motorist stops at the first signal but not at the second one.

c) P(at exactly one signal)

[tex]P(A\cap B') + P(A\cap 'B) = P(A\cup B) - P(A\cap B) \\P(A\cap B') + P(A\cap 'B) = 0.75 - 0.15 = 0.6[/tex]

0.6 is the probability that the motorist stops at exactly one signal.

Final answer:

To calculate various probabilities related to stopping at traffic signals, we use given values to determine a 15% chance of stopping at both signals, a 20% chance of stopping at the first but not the second, and a 60% chance of stopping at exactly one signal.

Explanation:

The question involves calculating probabilities of stopping at traffic signals. Given are the probabilities of stopping at the first (0.35) and second (0.55) signals, and the probability of stopping at at least one signal (0.75). Using these, we can find the probabilities for various scenarios.

A. Probability of stopping at both signals:

To find this, we use the formula: P(A and B) = P(A) + P(B) - P(A or B). Here, P(A or B) is the probability of stopping at least at one signal, which is given as 0.75. Thus, the calculation would be 0.35 + 0.55 - 0.75 = 0.15. So, there is a 15% chance of stopping at both signals.

B. Probability of stopping at the first signal but not the second one:

This can be calculated by subtracting the probability of stopping at both signals from the probability of stopping at the first signal: 0.35 - 0.15 = 0.20. Therefore, there is a 20% chance of stopping at the first signal but not the second.

C. Probability of stopping at exactly one signal:

This involves adding the probabilities of stopping only at the first signal or only at the second signal. We've already calculated the first part as 0.20. For the second part, subtract the probability of stopping at both signals from stopping at the second signal: 0.55 - 0.15 = 0.40. Adding these together, 0.20 + 0.40 = 0.60, there is a 60% chance of stopping at exactly one signal.

Answer:

(a) The proportion of students taking the MCAT had a score over 519 is 0.0367.

(b) The percentage of students who scored more than 519 in the MCAT is 3.67%.

Step-by-step explanation:

Assuming that the sample size or the number of students taking the MCAT in 2015 is large, the sampling distribution of scores follow a normal distribution.

Let X = score of a student

Given:

Mean = [tex]\mu=500[/tex]

Standard deviation = [tex]\sigma=10.6[/tex]

(a)

Compute the probability of students who scored more than 519 as follows:

[tex]P(X>519)=P(\frac{X-\mu}{\sigma}> \frac{519-500}{10.6})\\=P(Z>1.7925)\\=1-P(Z<1.7925)[/tex]

Use the z-table to determine the probability.

[tex]P(X>519)=P(\frac{X-\mu}{\sigma}> \frac{519-500}{10.6})\\=P(Z>1.7925)\\=1-P(Z<1.7925)\\=1-0.9633\\=0.0367[/tex]

Thus, the probability of students who scored more than 519 is 0.0367.

(b)

Convert the probability of students who scored more than 519 to percentage

[tex]=0.0367\times100\\=3.67\%[/tex]

Thus, the percentage of students who scored more than 519 is 3.67%.

Final answer:

The z-score for 519 is approximately 1.79, which correlates to about 3.67% of students scoring higher than 519.

Explanation:

To find the proportion of students scoring over 519 on the MCAT, first, we must calculate the z-score. The z-score tells us how many standard deviations an element is from the mean. Since the mean score is 500.0 and the standard deviation is 10.6, the z-score formula is z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation. For a score of 519, the z-score would be z = (519 - 500.0) / 10.6 ≈ 1.79. To find the proportion of students scoring above this z-score, we would look up the z-score in a standard normal distribution table (Table A), or use a statistical software to find that the area to the right of z=1.79. This area corresponds to the proportion of students scoring higher than 519. Since standard normal distribution tables are commonly used and we don't have one provided here, let's assume that the area to the right of z=1.79 is approximately 3.67%.

Answer:

a) 62

b) 24

Step-by-step explanation:

For A, add the students who watched only one movie: 18+24+20=62

For B, look at how many students only watched Star Wars: 24

Additional information to complete the question:

How long does it take for a short pulse of light to travel from one end of the glass to the other?

Express your answer in terms of some or all of the variables f and L. Use the numeric value given for n in the introduction.

T = ___________ s

Answer:

 [tex]T = \frac{15}{f}[/tex]

Step-by-step explanation:

Given:

Thickness og glass = L

Index of refraction n=1.5

Frequency = f

[tex]Wavelength = \frac{L}{10}[/tex]

λ(air) [tex]= \frac{L}{10}[/tex]

λ(glass) = λ(air) / n

             = [tex]\frac{\frac{L}{10}}{1.5}[/tex]

             = [tex]\frac{L}{10} * \frac{1}{1.5}[/tex]

             = [tex]\frac{L}{15}[/tex]

V(glass) = fλ(glass)

              [tex]= f * \frac{L}{15}[/tex]

          [tex]T = \frac{L}{V_{glass}} = \frac{15}{f}[/tex]

Answer:

48

Step-by-step explanation:

to find erea you just multiply one number by the other

Answer:48

Step-by-step explanation:

Answer:

2,136 yards

Step-by-step explanation:

Since the roads met at a right angle, the distance between Dale and Betty can be interpreted as the hypotenuse of a right triangle with sides measuring 2000 yd and 750 yd. The distance between them is:

[tex]d^2=2000^2+750^2\\d=\sqrt{2000^2+750^2}\\d=2,136\ yards[/tex]

Dale and Betty are 2,136 yards away from each other after two minutes.

The ratio of the radius of particle A to that of particle B in uniform circular motion is approximately 2.437.

To find the ratio of the radius of particle A to that of particle B in uniform circular motion, let's first consider the equations of motion for uniform circular motion. The centripetal acceleration, a_c, is given by the equation a_c = v^2/r, where v is the velocity and r is the radius of the circle. The period of motion, T, is the time it takes for one complete revolution around the circle. Given that the acceleration of particle A is 4.9 times that of particle B, we can write the equation a_A = 4.9 * a_B. Also, the period of particle B is 2.4 times the period of particle A, so we can write the equation T_B = 2.4 * T_A.

Next, we can use the equations of motion to express the velocity and period in terms of the acceleration and radius. From the equation a_c = v^2/r, we can rearrange it to solve for v: v = sqrt(a_c * r). By substituting this expression for v into the equation T = 2 * pi * r / v, we can solve for the period T in terms of a_c and r. Plugging these expressions for the velocities and periods of particles A and B into the equations a_A = 4.9 * a_B and T_B = 2.4 * T_A, we can form an equation that relates the radii of the two particles: sqrt(a_A * r_A) = 4.9 * sqrt(a_B * r_B) and 2 * pi * r_B / sqrt(a_B * r_B) = 2.4 * (2 * pi * r_A / sqrt(a_A * r_A)). Simplifying these equations, we can solve for the ratio of the radii r_A/r_B and find that it is approximately 2.437.

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To find the ratio of the radius of motion between particles A and B, we can use the equations of uniform circular motion and the given information. The ratio is approximately 2.21.

To find the ratio of the radius of motion between particles A and B, we need to analyze the given information. Let's denote the acceleration of particle B as aB and the acceleration of particle A as aA. We're told that aA is 4.9 times aB, and the period of particle B is 2.4 times the period of particle A.

From the equations of uniform circular motion, we know that the acceleration is given by a = (4π2)/T2, where T is the period. Since aA = 4.9aB and TB = 2.4TA, we can set up the following equation:

(4π2)/TA2 = 4.9(4π2)/TB2

After canceling out common terms, we'll find that (TA2) / (TB2) = 4.9. Taking the square root of both sides, we get TA / TB = √4.9 = 2.21. Since the period is inversely proportional to the angular velocity, we can conclude that the ratio of the radii of motion is approximately 2.21.

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Answer:

D. 3.35

Step-by-step explanation:

First we need to form an equation and solve it to find the number of weeks when the prices were the same. Because the prices were the same we can say that b = c, and therefore form the equation:

2.35 + 0.25x = 1.75 + 0.4x - Now we nee to solve it and find x.

2.35 - 1.75 = 0.4x - 0.25x

0.6 = 0.15x

x = 0.6 ÷ 0.15

x = 4 weeks

So now we substitute x into the equation for beef and find the price.

b = 2.35 + (0.25 × 4)

b = 2.35 + 1

b = $3.35 per pound

The price per pound of beef when it was equal to the price of chicken is $3.35 per pound.

The prices of beef and chicken are represented by the following linear equations :

Price of beef, B :

B = 2.35 + 0.25x

Price of chicken, C :

C = 1.75 + 0.40x

Where x in both equations represents x weeks after July 1 of last summer :

Firstly :

We find the week in which the price of beef and chicken are the same :

Beef = Chicken

2.35 + 0.25x = 1.75 + 0.40x

We solve for x

2.35 - 1.75 = 0.40x - 0.25x

0.60 = 0.15x

x = 0.60 / 0.15

x = 4

Therefore, 4 weeks after July 1 of last summer, the price of beef and chicken were the same.

Therefore, the price per pound of beef in the 4th week is :

B = 2.35 + 0.25(4)

B = 2.35 + 1

B = 3.35

The price per pound of beef when it was equal to the price of chicken is $3.35 per pound.

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Answer:

8 rooms

Step-by-step explanation:

39 - 7 = 32. 32/4 per room is 8 rooms.

Final answer:

The equation to find the number of backstage rooms needed is (Total volunteers - Stage volunteers) / Volunteers per room = Number of rooms. Using the values provided (39 - 7) / 4, we find that there were eight backstage rooms at the dance recital.

Explanation:

Ms. Deutsch needs seven parent volunteers to help students get on and off stage and four parent volunteers per backstage room. The total number of parent volunteers used on the day of the recital is 39. To find the number of backstage rooms, we subtract the seven parent volunteers required for stage assistance from the total, leaving us with 32 volunteers. We then divide this number by the four parent volunteers per room, resulting in eight backstage rooms.

Initial number of volunteers required for stage assistance: 7

Each room requires: 4 volunteers

Total volunteers: 39

Volunteers for rooms: 39-7 = 32

Number of rooms: 32 / 4 = 8

Answer:

Step-by-step explanation:

We find what number we multiply by another number to get 54, 24, 18

54: 1, 2, 3, 6, 9, 18, 27, 54

24: 1, 2, 3, 4, 6, 8, 12, 24

18: 1, 2, 3, 6, 9, 18

Now the numbers that repeat in these sets are the common factors

We got 1, 2, 3, and 6, which make these our common factors

Answer:1,2,3,6

Step-by-step explanation:

The factors of 54,24 and 18 are 1,2,3,6

Answer:

(a) 7.69%

(b) 25%

(c) 2.78%

Step-by-step explanation:

(a)

In a deck of 52 cards there are 4 aces.

The odds in favor or the probability of selecting an ace is:

[tex]P(Ace) = \frac{Number\ of\ aces}{Number\ of\ cards\ in\ total}\\ =\frac{4}{52}\\ =0.076923\\\approx7.69\%[/tex]

Thus, the probability of selecting an ace from a random deck of 562 cards is 7.69%.

(b)

The outcomes of each toss of a coin is independent of the other, since the result of the previous toss does not affect the result of the current toss.

The probability that both the tosses will end up in heads is:

[tex]P(2\ Heads)=P(1^{st}\ Head)\times P(2^{nd}\ Head)\\=\frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{4}\\ =0.25\ or\ 25\%\\[/tex]

Thus, the probability that both the tosses will end up in heads is 25%.

(c)

The sample space of two dice consists of 36 outcomes in total.

Out of these 36 outcomes there is only 1 Boxcar, i.e. two sixes.

The probability of a boxcar when two dice are rolled is:

[tex]P(Boxcar)=\frac{Favorable\ outcomes}{Total\ no.\ of\ outcomes}\\= \frac{1}{36}\\ =0.027777\\\approx2.78\%[/tex]

Thus, the probability of a boxcar when two dice are rolled is 2.78%.

Final answer:

To find the odds in favor of specific events in probability theory, one must compare the number of successful outcomes to the number of unsuccessful ones. For selecting an ace from a deck of cards, the odds are 1:12; for getting two heads from two coin tosses, the odds are 1:3; and for rolling two sixes with two dice, the odds are 1:35.

Explanation:

The question asks for the odds in favor of several different probabilistic events, which relate to the field of probability theory within mathematics. Here's how to calculate the odds for each of the requested scenarios:

(a) Odds in favor of a card being an ace: There are 4 aces in a standard 52-card deck. The odds in favor are the number of ways the event can occur (4 aces) to the number of ways the event can fail to occur (52 - 4 = 48 non-aces), which simplifies to 1:12.

(b) Odds in favor of two heads when a coin is tossed twice: The probability of getting a head on one coin toss is 1/2, and since the two tosses are independent, the probability of getting two heads is (1/2) * (1/2) = 1/4. The odds in favor are calculated by taking the probability of the event occurring (1 chance) against the probability of it not occurring (3 chances), which gives us odds of 1:3.

(c) Odds in favor of rolling boxcars (two sixes) with two dice: Each die has a 1/6 chance of rolling a six, so the probability of rolling two sixes is (1/6) * (1/6) = 1/36. The odds in favor are the number of successful outcomes (1) against the number of all other outcomes (35), resulting in odds of 1:35.

Answer:

1. (-∞,1) 2. (-∞,4]

Step-by-step explanation:

-2x-4 > -6

-2x > -2

x < 1

3(x+2) ≤ 18

3x+6 ≤ 18

3x ≤ 12

x ≤ 4

Answer:

a) [tex]\Omega=\{1, 2, 31, 32, 41, 42, 341, 342, 431, 432\}[/tex]

b) A={1, 2}

c) B={341, 342, 431, 432}

d) C={31, 32, 341, 342, 431, 432}

e) D={1, 31, 41, 341, 431}

f) E={41, 42, 341, 342, 431, 432}

Step-by-step explanation:

We know that four candidates are to be interviewed for a job. Two of them, numbered 1 and 2, are qualified, and the other two, numbered 3 and 4, are not.  

a) We get a set of all possible outcomes:

[tex]\Omega=\{1, 2, 31, 32, 41, 42, 341, 342, 431, 432\}[/tex]

b) Let A be the event that only one candidate is interviewed.

We get a set of all possible outcomes:

A={1, 2}

c) Let B be the event that three candidates are interviewed.

We get a set of all possible outcomes:

B={341, 342, 431, 432}

d)  Let C be the event that candidate 3 is interviewed.

We get a set of all possible outcomes:

C={31, 32, 341, 342, 431, 432}

e) Let D be the event that candidate 2 is not interviewed.

We get a set of all possible outcomes:

D={1, 31, 41, 341, 431}

f) Let E be the event that candidate 4 is interviewed.

We get a set of all possible outcomes:

E={41, 42, 341, 342, 431, 432}

We conclude that the events A and E are mutually exclusive.  

We conclude that the events B and E are not mutually exclusive.

We conclude that the events C and E are not mutually exclusive.

We conclude that the events D and E are not mutually exclusive.

Answer:

The balance will be $7,164.31.

Step-by-step explanation:

The compound interest formula is given by:

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

Where A is the amount of money, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit t and t is the time the money is invested or borrowed for.

In this problem, we have that:

[tex]P = 6000, r = 0.06[/tex]

Semi-annually is twice a year, so [tex]n = 2[/tex]

We want to find A when [tex]t = 3[/tex]

[tex]A = P(1 + \frac{r}{n})^{nt}[/tex]

[tex]A = 6000(1 + \frac{0.06}{2})^{2*3}[/tex]

[tex]A = 7164.31[/tex]

The balance will be $7,164.31.

Answer:

[tex]P(X<30)=P(\frac{X-\mu}{\sigma}<\frac{30-\mu}{\sigma})=P(Z<\frac{30-35}{10})=P(Z<-0.5)[/tex]

And we can find this probability using the normal standard table or excel and we got:

[tex]P(Z<-0.5)=0.309[/tex]

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the time spent for each customer of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(35,10)[/tex]  

Where [tex]\mu=35[/tex] and [tex]\sigma=10[/tex]

We are interested on this probability

[tex]P(X<30)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

[tex]P(X<30)=P(\frac{X-\mu}{\sigma}<\frac{30-\mu}{\sigma})=P(Z<\frac{30-35}{10})=P(Z<-0.5)[/tex]

And we can find this probability using the normal standard table or excel and we got:

[tex]P(Z<-0.5)=0.309[/tex]

And the excel code for this case would be : "=NORM.DIST(-0.5,0,1,TRUE)"

To calculate the probability that a customer spends less than 30 minutes in the store, we apply the Z score formula (Z = (X - μ)/σ), resulting in a Z score of -0.5. Using a Z-table, we find the corresponding probability to be approximately 30.85%.

This problem involves understanding the concept of a normal distribution, including the mean and standard deviation. The mean here is the average time customers spend in the store, which is 35 minutes, and the standard deviation is 10 minutes. The question is asking us to find the probability that a customer spends less than 30 minutes in the store.

We can calculate this using the concept of a Z score, which is given by the formula Z = (X - μ)/σ where X is the value we're interested in, μ is the mean, and σ is the standard deviation. Plugging in the values, we get Z = (30 - 35)/10 = -0.5. Looking this value up in a Z-table, we find that the probability a customer spends less than 30 minutes in the department store is approximately 0.3085, or 30.85%.

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Answer:

4 minutes

Step-by-step explanation:

There are two hours from 11:30 a.m. to 1:30 p.m

The hourly rate of service is:

[tex]r=\frac{300}{2}=150\ customers/hour[/tex]

If the average number of customers in the system (n) is 10, the time that a customer spends in process is given by:

[tex]t=\frac{n}{r} =\frac{10}{150}=0.06667\ hours[/tex]

Converting it to minutes:

[tex]t= 0.066667\ hours*\frac{60\ minutes}{1\ hour}\\t=4\ minutes[/tex]

A customer spends, on average, 4 minutes in process.

Answer:

30.56%

Step-by-step explanation:

Let the sides on each dice be labeled as 1, 2a, 2b, 2c, 3, 4.

The sample space for the sum of the values being 4 is:

S={1,3; 3,1; 2a,2a; 2a,2b; 2a,2c; 2b,2a; 2b,2b; 2b,2c; 2c,2a; 2c,2b; 2c,2c}

There are 11 possible sums out of the 36 possible outcomes that result in a sum of 4. Therefore, the probability their total is 4 is:

[tex]P(S) =\frac{11}{36}=0.3056 =30.56\%[/tex]

There is a 30.56% probability that their sum is 4.

Final answer:

The probability of getting a sum of 4 with two modified dice numbered 1,2,2,2,3,4 is 5/36. This is found by adding all possible combinations that total 4, considering the multiplicity of the number 2 on the dice.

Explanation:

To calculate the probability that the sum of two modified dice is 4, we must consider all possible combinations of rolls that could result in a total of 4. Each die is numbered with 1,2,2,2,3,4, so the outcomes that give us a sum of 4 are (1,3), (2,2), (3,1), and there are three different 2s on each die that can contribute to the sum.

Therefore, the probability of getting a sum of 4 with one die already showing 2 is the probability of rolling either a 1 or another 2 on the second die.

The total number of outcomes for one die is 6. To find the sum of 4:

This results in 1 + 3 + 1 = 5 favorable outcomes. Since there are a total of 36 possible outcomes when rolling two dice, the probability is 5/36.

Answer:

The probability that there will be more than one Phillips head screw = 0.1803 .

Step-by-step explanation:

We are given that there are 11 two-inch screws in a box of which 4 have a Phillips head and 7 have a regular head.

We are selecting 3 screws randomly from the box with replacement, so the probability that there will be more than one Phillips head screw is given by :

Now P(selecting 2 Phillips head screw with replacement) is given by :

 Selecting 2 Phillip head screw = [tex]\frac{4}{11}[/tex] *  [tex]\frac{4}{11}[/tex] = [tex]\frac{16}{121}[/tex]

      P(selecting three Phillips head screw) = [tex]\frac{4}{11}[/tex] *  [tex]\frac{4}{11}[/tex] *  [tex]\frac{4}{11}[/tex] = [tex]\frac{64}{1331}[/tex]

Therefore, Probability that there will be more than one Phillips head screw

                        =   [tex]\frac{16}{121}[/tex] + [tex]\frac{64}{1331}[/tex] = [tex]\frac{240}{1331}[/tex] = 0.1803 .

Answer:

The number of ways to select 5 diamonds and 3 clubs is 368,082.

Step-by-step explanation:

In a standard deck of 52 cards there are 4 suits each consisting of 13 cards.

Compute the probability of selecting 5 diamonds and 3 clubs as follows:

The number of ways of selecting 0 cards from 13 hearts is:

[tex]{13\choose 0}=\frac{13!}{0!\times(13-0)!} =\frac{13!}{13!}=1[/tex]

The number of ways of selecting 3 cards from 13 clubs is:

[tex]{13\choose 3}=\frac{13!}{3!\times(13-3)!} =\frac{13!}{13!\times10!}=286[/tex]

The number of ways of selecting 5 cards from 13 diamonds is:

[tex]{13\choose 5}=\frac{13!}{5!\times(13-5)!} =\frac{13!}{13!\times8!}=1287[/tex]

The number of ways of selecting 0 cards from 13 spades is:

[tex]{13\choose 0}=\frac{13!}{0!\times(13-0)!} =\frac{13!}{13!}=1[/tex]

Compute the number of ways to select 5 diamonds and 3 clubs as:

[tex]{13\choose0}\times{13\choose3}\times{13\choose5}\times{13\choose0} = 1\times286\times1287\times1=368082[/tex]

Thus, the number of ways to select 5 diamonds and 3 clubs is 368,082.