Answer:
D
Step-by-step explanation:
The population consists of all characteristics of interest and sample is portion or a subset of population. Here, 1502 adults 18 years or older are select ted from a population of all adults 18 years or older, so, 1502 adults are the sample. The measurement taken from sample is termed as statistic. The given value 69% is computed from a sample and thus it is a sample statistic.
The XO Group Inc. conducted a survey of 13,000 brides and grooms married in the United States and found that the average cost of a wedding is $29,858 (XO Group website, January 5, 2015). Assume that the cost of a wedding is normally distributed with a mean of $29,858 and a standard deviation of $5600.a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?
Answer:
a) 0.0392
b) 0.4688
c) At least $39,070 to be among the 5% most expensive.
Step-by-step explanation:
Problems of normally distributed samples can be 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 = 29858, \sigma = 5600[/tex]
a. What is the probability that a wedding costs less than $20,000 (to 4 decimals)?
This is the pvalue of Z when X = 20000. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20000 - 29858}{5600}[/tex]
[tex]Z = -1.76[/tex]
[tex]Z = -1.76[/tex] has a pvalue of 0.0392.
So this probability is 0.0392.
b. What is the probability that a wedding costs between $20,000 and $30,000 (to 4 decimals)?
This is the pvalue of Z when X = 30000 subtracted by the pvalue of Z when X = 20000.
X = 30000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30000 - 29858}{5600}[/tex]
[tex]Z = 0.02[/tex]
[tex]Z = 0.02[/tex] has a pvalue of 0.5080.
X = 20000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20000 - 29858}{5600}[/tex]
[tex]Z = -1.76[/tex]
[tex]Z = -1.76[/tex] has a pvalue of 0.0392.
So this probability is 0.5080 - 0.0392 = 0.4688
c. For a wedding to be among the 5% most expensive, how much would it have to cost (to the nearest whole number)?
This is the value of X when Z has a pvalue of 0.95. So this is X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 29858}{5600}[/tex]
[tex]X - 29858 = 5600*1.645[/tex]
[tex]X = 39070[/tex]
The wedding would have to cost at least $39,070 to be among the 5% most expensive.
Binomial Distribution. Surveys repeatedly show that about 40% of adults in the U.S. indicate that if they only had one child, they would prefer it to be a boy. Suppose we took a random sample of 15 adults and the number who indicated they preferred a boy was 8. This would be considered a rare event because the probability of 8 or more is so low.
True/False
Answer:
False
Step-by-step explanation:
We are given the following information:
We treat adult who prefer one child to be a boy as a success.
P(prefer one child to be a boy) = 40% = 0.4
Then the number of adults follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 15 and x = 8
We have to evaluate:
[tex]P(x \geq 8)\\= P(x = 8) + P(x = 9)+...+ P(x = 14) + P(x =15)\\\\= \binom{15}{8}(0.4)^{8}(1-0.4)^{7} +\binom{15}{9}(0.4)^{9}(1-0.4)^{6}+...\\\\...+\binom{15}{14}(0.4)^{14}(1-0.4)^{1} +\binom{15}{8}(0.4)^{15}(1-0.4)^{0}\\\\= 0.2131[/tex]
Since the probability of 8 or more is 0.2131 is not very small, thus, it is not a rare event.
Thus, the given statement is false.
Explain what a P-value is. What is the criterion for rejecting the null hypothesis using the P-value approach?
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.
Explanation: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.
The length of a Texas Pee Wee football field is 218 feet greater than its width. The area of the field is 20,160 square feet
Please show work !!
Answer: [tex]9.616 ft[/tex]
Step-by-step explanation:
The last part of the question is: Find the value of the width
If the Texas Pee Wee football field has a rectangular shape (as shown in the figure), where the width is [tex]w[/tex] and the length is [tex]218 ft w[/tex]; its area [tex]A[/tex] is:
[tex]A=20,160 ft^{2}=(length)(width)[/tex]
[tex]20,160 ft^{2}=(218 w)(w)[/tex]
Isolating [tex]w[/tex]:
[tex]w=\sqrt{\frac{20,160 ft^{2}}{218}}[/tex]
Finally:
[tex]w=9.616 ft[/tex] This is the widht of the Texas Pee Wee football field
The Texas Pee Wee football field has a width of approximately 70 feet and a length of 288 feet, calculated by solving a quadratic equation derived from the given area and the relationship between length and width.
Step-by-Step Explanation:
Let the width of the field be denoted as w.
Therefore, the length of the field is w + 218 feet.
The area of the rectangle (football field) is given by the formula:
Area = length x width.
Substituting the given values into the formula, we have:
20,160 = w × (w + 218).
This results in a quadratic equation:
w² + 218w - 20,160 = 0.
We will solve this quadratic equation using the quadratic formula:
w = (-b ± √(b² - 4ac)) / 2a, where a = 1, b = 218, and c = -20,160.
Calculate the discriminant:
b² - 4ac = 218² - 4 × 1 × (-20,160) = 47524 + 80640 = 128164.
Find the square root of the discriminant:
√128164 ≈ 357.94.
Substitute back into the quadratic formula:
w = (-218 ± 357.94) / 2.
This results in two potential solutions:
w = (357.94 - 218) / 2 ≈ 69.97 (approximately 70 feet) and w = (-218 - 357.94) / 2 (a negative value, which is not possible for width).
Thus, the width w is approximately 70 feet.
Substitute the width back into the length formula:
length = 70 + 218 = 288 feet.
Therefore, the dimensions of the Texas Pee Wee football field are approximately 70 feet in width and 288 feet in length.
You have two fair, six-sides dice. However, the dice have been modified so that instead or 1,2,3,4,5,6 the sides are numbered 1,2,2,2,3,4. (Write all answers as fractions, not decimals) When the two dice are thrown, what is the probability their total is 4
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:
(1,3) - There is 1 way to roll a 1 and 1 way to roll a 3.(2,2) - There are 3 ways to roll a 2 on the first die and 3 ways to roll a 2 on the second die, but since the outcome is indistinguishable (2,2) is considered once, making it 3 ways in total.(3,1) - There is 1 way to roll a 3 and 1 way to roll a 1.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.
2.65 Consider the situation of Exercise 2.64. Let A be the event that the component fails a particular test and B be the event that the component displays strain but does not actually fail. Event A occurs with probability 0.20, and event B occurs with probability 0.35. (a) What is the probability that the component does not fail the test? (b) What is the probability that the component works perfectly well (i.e., neither displays strain nor fails the test)? (c) What is the probability that the component either fails or shows strain in the test?
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.
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A survey of 400 non-fatal accidents showed that 109 involved the use of a cell phone. Construct a 99% confidence interval for the proportion of fatal accidents that involved the use of a cell phone
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).
The following histogram presents the amounts of silver (in parts per million) found in a sample of rocks. One rectangle from the histogram is missing. What is its height?
Answer:
The height of the missing rectangle is 0.15Explanation:
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:
0.2 + 0.3 + 0.15 + 0.1 + 0.1 = 0.852. Missing frequency:
1 - 0.85 = 0.153. Conclusion:
The height of the missing rectangle is 0.15The 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. A. What is the probability that the diameter of a dot exceeds 0.0026 inch? B. What is the probability that a diameter is between 0.0014 and 0.0026? C. What standard deviation of diameters is needed so that the probability in part (b) is 0.995?
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.
A. The probability that the diameter of a dot exceeds 0.0026 inch is 0.0668B. The probability that the diameter is between 0.0014 and 0.0026 inch is 0.8664C. The standard deviation needed for a probability of 0.995 is 2.135 × 10⁻⁴Reasons:
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
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Simplify 25 ^1/2 using the radical form
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.
Explanation: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.
A 200-liter tank initially full of water develops a leak at the bottom. Given that 30% of the water leaks out in the first 5 minutes, find the amount of water left in the tank 10 minutes after the leak develops if the water drains off at a rate that is proportional to the amount of water present.
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.
Explanation: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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Find the volume of a cube with side length of 7 in.
147
343
49
215
Answer:
48
Step-by-step explanation:
to find erea you just multiply one number by the other
Answer:48
Step-by-step explanation:
b=2.35 + 0.25x
c=1.75+0.40x
In the equations above,b and c represent the price per pound,in dollars, of beef and chicken,respectively,x weeks after July 1 of last summer.What was the price per pound of beef when it was equal to the price per pound of chicken?
A.2.60
B.2.85
C.2.95
D.3.35
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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A survey reveals that each customer spends an average of 35 minutes with a standard deviation of 10 minutes in a department store. Assuming the distribution is normal, what is the probability a customer spends less than 30 minutes in the department store?
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%.
Explanation: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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The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is 0.35, the analogous probability for the second signal is 0.55, and the probability that he must stop at at least one of the two signals is 0.75.
What is the probability that he must stop:
A. at both signals?
B. at the first signal but not at the second one?
C. at exactly one signal?
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.
You have 200 dice in a bag. One of the dice has a six on all sides so it will land on a six every time you roll it. The other 199 are normal dice with six sides, each with a different number. You randomly pick one of the dice from the bag and roll it three times. It lands on six all three times. What is the probability it is the die that always lands on six and what is the probability it is a normal die?
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]
Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 4.9 times that of particle B. The period of particle B is 2.4 times the period of particle A. The ratio of the radius of the motion of particle A to that of particle B is closest to
The ratio of the radius of particle A to that of particle B in uniform circular motion is approximately 2.437.
Explanation: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.
Explanation: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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The perimeter of a rhombus is 64 and one of its angles has measure 120. Find the lengths of the diagonals.
Answer:
8[tex]\sqrt{2}[/tex]
Step-by-step explanation:
64/4=16
so that is 16 on each side.
The diagonal of the rhombus create a right triangle.
We then use the Pythagorean theorem.
[tex]a^{2} +b^{2} =c^{2}[/tex]
[tex]16^{2} +16^{2} =c^{2}[/tex]
[tex]256+256=c^{2}[/tex]
[tex]512 =c^{2}[/tex]
[tex]\sqrt{512} =c[/tex]
8[tex]\sqrt{2}[/tex]
The length of the diagonals of a rhombus with perimeter of 64 and one of its angles as 120 degrees are 16 units and 27.71 units
Properties of a rhombusThe diagonals are angle bisectorsThe 4 sides are congruent.The diagonal are perpendicular bisectors
Therefore,
perimeter = 4l
where
l = length
64 = 4l
l = 64 / 4
length = 16
One of its angle is 120°. Therefore, let's use the angle to find the length of the diagonal.
Using trigonometric ratio,
cos 60° = adjacent / hypotenuse
cos 60° = x / 16
x = 16 × cos 60
x = 8
2(x) = diagonal
diagonal = 16 units
The second diagonal length
sin 60° = opposite / hypotenuse
sin 60 = y / 16
y = 16 × sin 60
y = 13.8564064606
y = 13.85
Therefore,
diagonal = 2(13.85) = 27.71 units
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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. The candidates are interviewed at random, and the first qualified candidate interviewed will be hired. The outcomes are the sequences of candidates that are interviewed. So one outcome is 2, and another is 431.
a. List all the possible outcomes.
b. Let A be the event that only one candidate is interviewed. List the outcomes in A.
c. Let B be the event that three candidates are interviewed. List the outcomes in B.
d. Let C be the event that candidate 3 is interviewed. List the outcomes in C.
e. Let D be the event that candidate 2 is not interviewed. List the outcomes in D
f. Let E be the event that candidate 4 is interviewed. Are A and E mutually exclusive? How about B and E, C and E, D and E?
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.
what are the common factors for 54,24,18
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
At the dance recital, Ms. Deutsch needs seven parent volunteers to help students get on and off stage, plus four parent volunteers per room of students. On the day of the recital, Ms. Deutsch uses 39 parent volunteers. How many backstage rooms were there?
plz help Like now plz
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
Dale and Betty go through a traffic light at the same time but Dale goes straight and Betty turns right. After two minutes Dale is 2000 yd from the intersection and Bettyis 750 yd from the intersection. Assuming the roads met at a right angle and both were perfectly straight, how far are Dale and Betty away from each other after two minutes?
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.
Solve, graph, and give interval notation for the compound inequality:
−2x − 4 > −6 AND 3(x + 2) ≤ 18
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
When Alice spends the day with the babysitter, there is a 0.6 probability that she turns on the TV and watches a show. Her little sister Betty cannot turn the TV on by herself. But once the TV is on, Betty watches with probability 0.8. Tomorrow the girls spend the day with the babysitter.a) What is the probability that both Alice and Betty watch TV tomorrow?b) What is the probability that Betty watches TV tomorrow?c) What is the probability that only Alice watches TV tomorrow?
Answer:
a) There is a 48% probability that both Alice and Betty watch TV tomorrow.
b) There is a 48% probability that Betty watches TV tomorrow.
c) There is a 12% probability that only Alice watches TV tomorrow.
Step-by-step explanation:
We have these following probabilities:
A 60% probability that Alice watches TV.
If Alice watches TV, an 80% probability Betty watches TV.
If Alice does not watch TV, a 0% probability that Betty watches TV, since she cannot turn the TV on by herself.
a) What is the probability that both Alice and Betty watch TV tomorrow?
Alice watches 60% of the time. Betty watches in 80% of the time Alice watches. So:
[tex]P = 0.6*0.8 = 0.48[/tex]
There is a 48% probability that both Alice and Betty watch TV tomorrow.
b) What is the probability that Betty watches TV tomorrow?
Since Betty only watches when Alice watches(80% of the time), this probability is the same as the probability of both of them watching. So
[tex]P = 0.6*0.8 = 0.48[/tex]
There is a 48% probability that Betty watches TV tomorrow.
c) What is the probability that only Alice watches TV tomorrow?
There is a 60% probability that Alice watches TV tomorrow. If she watches, there is an 80% probability that Betty watches and a 20% probability she does not watch.
So
[tex]P = 0.6*0.2 = 0.12[/tex]
There is a 12% probability that only Alice watches TV tomorrow.
Assume that a cross is made between a heterozygous tall pea plant and a homozygous short pea plant. Fifty offspring are produced in the following frequency:30 = tall20 = shortNull hypothesis: The deviations from a 1:1 ratio (25 tall and 25 short) are due to chance.What is the Chi-square value associated with the appropriate test of significance?
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.
A beam of light from a monochromatic laser shines into a piece of glass. The glass has thickness L and index of refraction n=1.5. The wavelength of the laser light in vacuum is L/10 and its frequency is f. In this problem, neither the constant c nor its numerical value should appear in any of your answers.
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]
Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). A new version of the exam was introduced in spring 2015 and is intended to shift the focus from what applicants know to how well they can use what they know. One result of the change is that the scale on which the exam is graded has been modified, with the total score of the four sections on the test ranging from 472 to 528 . In spring 2015 the mean score was 500.0 with a standard deviation of 10.6 . Use Table A to find the answers to the two questions.
(a) What proportion of students taking the MCAT had a score over 519 ?
(b) Compute the proportion and then enter the answer as a percentage rounded to two decimal places.
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%.
Find the probability for the experiment of tossing a coin three times. Use the sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.
1. The probability of getting exactly one tail
2. The probability of getting exactly two tails
3. The probability of getting a head on the first toss
4. The probability of getting a tail on the last toss
5. The probability of getting at least one head
6. The probability of getting at least two heads
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]
Find the balance of $6,000 deposited at 6% compounded semi-annually for 3 years
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.
Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler test). Find the probability that a randomly selected adult has an IQ greater than 131.5. Group of answer choices
Answer:
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 100
Standard Deviation, σ = 15
We are given that the distribution of IQ score is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(IQ greater than 131.5)
P(x > 131.5)
[tex]P( x > 131.5) = P( z > \displaystyle\frac{131.5 - 100}{15}) = P(z > 2.1)[/tex]
[tex]= 1 - P(z \leq 2.1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 131.5) = 1 - 0.982 = 0.018 = 1.8\%[/tex]
0.018 is the probability that a randomly selected adult has an IQ greater than 131.5