Answer:
The answer is $1000.
Step-by-step explanation:
The policy was issued with a $500 deductible and a limit of four deductibles per calendar year.
As given that two claims were paid in September 2013, each incurring medical expenses in excess of the deductible.
So, the family's out-of-pocket medical expenses for 2013 will be :
[tex]500+500=1000[/tex] dollars
As the limit was up to 4 deductibles in a calendar year, and in 2013, there were 2 claims, so that sums up to be $1000.
The family's out-of-pocket medical expenses for 2013 would be $1000, as they paid the $500 deductible for each of the two claims made that year, with their health insurance policy limiting to four deductibles per year.
Explanation:The subject of the question involves calculating the out-of-pocket medical expenses for a family under their health insurance policy, which includes understanding how deductibles work. In the scenario given, the family purchased a policy with a $500 deductible and a limit of four deductibles per calendar year. In 2013, they made two claims where each exceeded the deductible amount. Therefore, their out-of-pocket expenses for 2013 would be two times the deductible amount, since the policy has a limit of four deductibles per year but only two claims were filed and paid within that year.
Mathematically, this can be calculated as:
Claim 1 in September 2013: $500 (deductible)Claim 2 in September 2013: $500 (deductible)Total out-of-pocket expenses for 2013: $500 + $500 = $1000.
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Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
A. 1/24
B. 1/8
C. 1/4
D. 1/3
E. 3/8
Answer:
The probability that only 1 letter will be put into the envelope with its correct address is [tex]\frac{1}{3}[/tex]
Step-by-step explanation:
Given:
Number of Letters=4
Number of addresses= 4
To Find:
The probability that only 1 letter will be put into the envelope with its correct address=?
Solution:
Let us assume first letter goes in correct envelope and others go in wrong envelopes, then
=> Probability putting the first letter in correct envelope =[tex]\frac{1}{4}[/tex]
=> Probability putting the second letter in correct envelope =[tex]\frac{2}{3}[/tex]
=> Probability putting the third letter in correct envelope= [tex]\frac{1}{2}[/tex]
=> Probability putting the fourth letter in correct envelope = 1;
( only 1 wrong addressed envelope is left);
This event can occur with other 3 envelopes too.
Hence total prob. = [tex]4\times(\frac{1}{4}\times\frac{2}{3}\times\frac{1}{2}\times1)[/tex]
=> [tex]\frac{1}{3}[/tex]
Hello, I'm posting this again because I really need help, I will give brainliest, I'm pretty desperate. ;--;
What is the inequality represented by the graph?
Thank you !!
Answer:
[tex]y < - \times + 2[/tex]
and
[tex]y \geqslant 2x + 4[/tex]
Step-by-step explanation:
Dotted line means regular < and >
Solid line mean (< or equal to) and > (or equal to)
They're asking for an equation for both lines, which you can use the formula y = mx + b, but in this case you'll be using y < or y > since it's an inequality.
The m in the equation represents the slopeThe b in the equation represents where the line intercepts on the y-axisFor the dotted line: Its dotted so you already know it's a regular sign (< and >). We have to find the slope of the dotted line, which is m. The formula for m = (y2 - y1) ÷ (x2 - x1), which means you choose two points that the dotted line intercepts with. (0, 2) and (2,0) are two points the line goes through. Now plug it into the slope formula. (0 - 2) ÷ (2 - 0) = -2/2 = -1
The line intercepts at 2 on the y-axis and the area below the dotted line is shaded. When it's shaded below, the sign is < therefore y < -1x + 2
For the solid line: Its solid so the sign in underlined indicating equal to or (</>). Do the exact same thing you did for the dotted line. Slope formula and where the line intercepts the y-axis. Let's do (-2,0) and (0,4), then (4 - 0) ÷ (0 - (-2)) = 4/2 = 2.
The line intercepts at 4 on the y-axis and the area above the solid line is shaded. When its shaded above, the sign is > therefore y > or equal to (underline it) 2x + 4
Complete the proof.
Prove: △FIJ ≅ △HGJ
A) AAS
B) ASA
C) SAS
D) SSS
Answer:
The answer is SSS.
Step-by-step explanation:
It is proved that △FIJ ≅ △HGJ By Side side Side Congruence Property.
Thus the correct option is D.
What is the congruent triangle?Two triangles are said to be congruent if the length of the sides is equal, a measure of the angles are equal and they can be superimposed.
Given:
In △FIJ and △HGJ
Segment FI ≅ segment GH
Segment FJ = segment HJ (by definition of midpoint)
Segment GJ= segment IJ (by definition of midpoint)
∴ By Side side Side Congruence Property
△FIJ ≅ △HGJ by SSS
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Find the point P on the line yequals=22x that is closest to the point (10,0 )(10,0).
What is the least distance between P and (10,0 )(10,0)?
There is no Graph to go in response to this question
Grant thinks he has reached the point where he has maximized his profit. However, because this can be tricky to determine, he is not sure. He decides to test it by selling one more unit. If Grant is correct in his assumption,
what should happen when he sells this additional unit?
a. Grant will break even on his sales
b. Profit will increase but at a slower rate
c. Marginal revenue will exceed marginal cost
d. Marginal cost will exceed marginal revenue
e. Profits will be reduced to zero
Answer:c
Step-by-step explanation:
Profit maximization happens with marginal revenue is equal to marginal cost, so if Grant's assumption was right before selling the extra unit, when he actually sells the extra unit, this will increase his revenue
That's why the answer is c
Marginal revenue will exceed marginal cost
An adult can lose or gain two pounds of water in the course of a day. Assume that the changes in water weight are uniformly distributed between minus two and plus two pounds in a day. What is the standard deviation of your weight over a day?
Answer: 1.15 pounds
Step-by-step explanation:
For uniform distribution.
The standard deviation is :
[tex]\sigma=\sqrt{\dfrac{(b-a)^2}{12}}[/tex]
, where a = Lower limit of interval [a,b].
b = Upper limit of interval [a,b].
Given : The changes in water weight are uniformly distributed between minus two and plus two pounds in a day.
i.e. Interval = [-2 , +2]
Here , a= -2 and b= 2
Then, the standard deviation is :
[tex]\sigma=\sqrt{\dfrac{(2-(-2))^2}{12}}[/tex]
[tex]\sigma=\sqrt{\dfrac{(2+2)^2}{12}}[/tex]
[tex]\sigma=\sqrt{\dfrac{16}{12}}=\sqrt{1.3333}=1.15468610453\approx1.15[/tex]
Hence, the standard deviation of your weight over a day = 1.15 pounds
The standard deviation of the uniform distribution representing an adult's daily change in weight due to water is around 1.155 pounds.
Explanation:The question is about the standard deviation of the adult weight changes due to gain or loss in water content which is uniformly distributed between minus two and plus two pounds in a day.
To calculate the standard deviation for this uniform distribution, you need to follow these steps:
The range of the distribution is the difference between the highest and lowest values. In this case, the range is 4 pounds (2 pounds of gain - (-2 pounds of loss)). The formula of standard deviation for a uniform distribution is: sqrt((range^2) / 12). Substituting the values, the answer would be sqrt((4^2) / 12), which equals to 1.155 pounds.
So, the standard deviation of your weight changes over a day due to the water flux is approximately 1.155 pounds.
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A spotlight is made by placing a strong light bulb inside a reflective paraboloid formed by rotating the parabola x^2=4y around its axis of symmetry (assume that x and y are in units of inches). In order to have the brightest, most concentrated light beam, how far from the vertex should the bulb be placed? Express your answer as a fraction or a decimal rounded to two decimal places.
Answer:
1.00 inches
Step-by-step explanation:
The distance from vertex to focus is "p" in the quadratic equation ...
x^2 = 4py
In the given equation, p=1. Since units are inches, ...
the bulb should be placed 1.00 inches from the vertex.
5*5 what si the answer
Answer:
25
Step-by-step explanation:
Quart cartons of milk should contain at least 32 ounces. A sample of 22 cartons was taken and amount of milk in ounces was recorded. We would like to determine if there is sufficient evidence exist to conclude the mean amount of milk in cartons is less than 32 ounces? a) Two sample t testb) One sample t testc) Two sample p testd) Matched pairs
Answer:
32 oz
Step-by-step explanation:
On New Year's Eve, the probability of a person driving while intoxicated or having an accident is 0.37. If the probability of driving while intoxicated is 0.32 and the probability of having a driving accident is 0.12, find the probability of a person having a driving accident while intoxicated
Answer:
The probability of a person having a driving accident while intoxicated is 0.07
Step-by-step explanation:
Hi, well, let´s put this on a formula, I think it is the best way to explain it.
[tex]P(A+I)=P(A)+P(I)-P(AorI)[/tex]
Where:
P(A+I) = Probability of having a driving accident while intoxicated.
P(A) = Probability of a person of having an accident.
P(I) = Probablity person being intoxicated.
P(A or I) = Probability of a person for being intoxicated or having an accident.
Therefore, things should look like this:
[tex]P(A+I)=0.12+0.32-0.37=0.07[/tex]
So, the probability of a person having a driving accident while intoxicated is 0.07.
Best of luck.
The probability of a person having a driving accident while intoxicated is 0.375 or 37.5%.
Explanation:To find the probability of a person having a driving accident while intoxicated, we can use the formula for conditional probability: P(A|B) = P(A and B) / P(B). In this case, A represents the event of having a driving accident and B represents the event of driving while intoxicated. The probability of driving while intoxicated is given as 0.32, and the probability of having a driving accident is given as 0.12. So, P(A and B) = 0.12 and P(B) = 0.32. Plugging these values into the formula, we get P(A|B) = 0.12 / 0.32 = 0.375. Therefore, the probability of a person having a driving accident while intoxicated is 0.375 or 37.5%.
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First one digit is chosen uniformly at random from f1; 2; 3; 4; 5g and is removed from the set; then a second digit is chosen uniformly at random from the remaining digits. What is the probability that an odd digit is picked the second time?
Answer:
[tex]\frac{3}{5}[/tex]
Step-by-step explanation:
Probability of choosing an odd number in the second turn is the sum of probabilities of choosing an odd number in second turn given that an odd number or an even number is picked in first turn.
Probability of getting an odd number in the first turn out of 1,2,3,4,5 is [tex]\frac{3}{5}[/tex]
Probability of getting an even number in the first turn out of 1,2,3,4,5 is [tex]\frac{2}{5}[/tex]
Probability of getting an odd number in second turn given that an odd number was picked in the first turn (remaining : 2 odd numbers out of 4) is [tex]\frac{1}{2}[/tex]
Probability of getting an odd number in second turn given that an even number was picked in the first turn (remaining : 3 odd numbers out of 4) is [tex]\frac{3}{4}[/tex]
Total probability is [tex]\frac{3}{5} \times \frac{1}{2} + \frac{2}{5} \times \frac{3}{4} = \frac{3}{5}[/tex]
The probability of choosing an odd digit in the second draw, considering all scenarios of the first draw, is 0.625.
Explanation:The student's question pertains to probability in a sequential selection scenario. It involves two sequential selections of digits from a certain set, specifically looking at the situation where an odd digit is selected in the second draw.
To address this, we first acknowledge that there are 5 digits to choose from initially: 1, 2, 3, 4, 5. However, once a digit is chosen and removed, 4 digits remain in the set for the second round of choosing. Among the remaining 4 digits, either two or three of them will be odd, depending on the parity (evenness or oddness) of the first digit chosen.
If an even digit is chosen first, three odd digits (1,3,5) will be left, thus the probability of choosing an odd digit the second time is 3 out of 4, or 0.75. If an odd digit is chosen first, two odd digits will be left, and the probability of choosing an odd digit in the second draw is then 2 out of 4, or 0.5. Finally, we consider the total probability over all possible first draws, yielding (1/2)*0.75 + (1/2)*0.5 = 0.625.
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A scientist estimates that the mean nitrogen dioxide level in a city is greater than 30 parts per billion. To test this estimate, you determine the nitrogen dioxide levels for 31 randomly selected days. The results (in parts per billion) are listed below.
Answer:
Step-by-step explanation:
incomplete. no results listed below
Can relative maximums and minimums be at the endpoints? I know global max/min can be at endpoints but I'm not sure about relative max/min.
Answer:
No
Step-by-step explanation:
Just because the derivative is 0 at a point doesn't necessarily mean it is a relative minimum or maximum. You must be able to evaluate the derivative on both sides of the point to determine if it changes signs. Since endpoints have only one side, they cannot be relative maximums or minimums.
Jimmy walks 9 miles West and 12 miles North. How much shorter is the diagonal distance from point A to point B than walking the distance of both sides?
Answer:
6 miles shorter
Step-by-step explanation:
Right now, Jimmy walked 21 miles. If he had gone diagonally, he would've walked only 15 miles. This is 6 miles shorter than before.
| The ultracentrifuge is an important tool for separating and analyzing proteins. Because of the enormous centripetal accelerations, the centrifuge must be carefully balanced, with each sample matched by a sample of identical mass on the opposite side. Any difference in the masses of opposing samples creates a net force on the shaft of the rotor, potentially leading to a catastrophic failure of the apparatus. Suppose a scientist makes a slight error in sample preparation and one sample has a mass 10 mg larger than the opposing sample. If the samples are 12 cm from the axis of the rotor and the ultracentrifuge spins at 70,000 rpm, what is the magnitude of the net force on the rotor due to the unbalanced samples
The net force on the rotor due to the unbalanced samples : 64.4 N
Further explanationCentripetal force is a force acting on objects that move in a circle in the direction toward the center of the circle
[tex]\large{\boxed{\bold{F= \frac{mv^2}{R}}}[/tex]
F = centripetal force , N
m = mass , Kg
v = linear velocity , m/s
r = radius , m
The speed that is in the direction of the circle is called linear velocity
Can be formulated:
[tex]\displaysyle v=2\pi.r.f[/tex]
r = circle radius
f = rotation per second (RPS)
The sample has a mass of 10 mg larger than the opposing sample. If the samples are 12 cm from the axis of the rotor and the ultracentrifuge spins at 70,000 rpm
Known
RPM = 70,000, convert to RPS = 70,000: 60 = 1166.6
r = 12 cm = 0.12 m
m = 10 mg = 10⁻⁵ kg
then
Linear velocity :
v = 2π.r.f[tex]\displaystyle v=2\times 3.14\times 0.12\times 1166.6\\\\v=879.15\:m/s[/tex]
Centripetal force :
[tex]\displaystyle F=\frac{10^{-5}\times (879.15)^2}{0.12}\\\\F=\boxed{\bold{64.4\:N}}[/tex]
Learn morethe average velocity
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resultant velocity
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velocity position
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Keywords: ultracentrifuge, samples, Centripetal force, linear velocity
The magnitude of the net force on the rotor due to the unbalanced samples is 64.4 Newton.
How to calculate the net force magnitude?From the information given, the velocity will be calculated as:
= 2πrf.
where, r = radius = 0.12
f = rotation per second = 70000/60 = 1166.6
Velocity will be:
= 2 × 3.14 × 0.12 × 1166.6
= 879.15 m/s
Therefore, the centripetal force will be:
= [10^-5 × (879.15)²] ) 0.12
= 64.4N
In conclusion, the magnitude of the net force is 64.4 Newton.
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Flying against the wind, an airplane travels 5760 kilometers in 6 hours. Flying with the wind, the same plane travels 6300 kilometers in 5 hours. What is the rate of the plane in still air and what is the rate of the wind?
Step-by-step explanation:
Let the speed of plane be p and speed of wind be w.
Flying against the wind, an airplane travels 5760 kilometers in 6 hours.
Here
Speed = (p-w) kmph
Time = 6 hours
Distance = 5760 kmph
Distance = Speed x Time
5760 = (p-w) x 6
p-w = 960 -----eqn 1
Flying with the wind, the same plane travels 6300 kilometers in 5 hours.
Here
Speed = (p+w) kmph
Time = 5 hours
Distance = 6300 kmph
Distance = Speed x Time
6300 = (p+w) x 5
p+w = 1260 -----eqn 2
eqn 1 + eqn 2
p-w + p +w = 960 + 1260
2p = 2220
p = 1110 kmph
Substituting in eqn 2
1110 + w = 1260
w = 150 kmph
Speed of plane = 1110 kmph
Speed of wind = 150 kmph
Two new drugs are to be tested using a group of 60 laboratory mice, each tagged with a number for identification purposes. Drug A is to be given to 22 mice, drug B is to be given to another 22 mice, and the remaining 16 mice are to be used as controls. How many ways can the assignment of treatments to mice be made? (A single assignment involves specifying the treatment for each mouse—whether drug A, drug B, or no drug.) (Enter the exact number or an equivalent algebraic expression.)
Answer:
[tex]\frac{60!}{22!22!16!}[/tex]
Step-by-step explanation:
As given, drug A is to be given to 22 mice, drug B is to be given to another 22 mice, and the remaining 16 mice are to be used as controls.
Required number of ways = Number of ways to select mice that gets drug A x the number of ways for mice that gets drug B x the number of ways the mice gets no drugs.
= [tex]\frac{60!}{22!38!} \times \frac{38!}{22!16!} \times1[/tex]
Solving this we get;
= [tex]\frac{60!}{22!22!16!}[/tex]
= 314,790,828,599,338,321,972,833,000
The top and bottom margins of a poster are 8 cm and the side margins are each 6 cm. If the area of printed material on the poster is fixed at 390 square centimeters, find the dimensions of the poster with the smallest area.
The top and bottom margins of a poster are 8 cm an
Width = (include units)
Height = (include units)
Final answer:
To find the dimensions of the poster with the smallest area, subtract the margins from the total dimensions of the poster. Set up an equation using the area of the printed material and find the dimensions that result in the smallest area. By substituting different values into the equation, the dimensions are approximately 30 cm by 56 cm.
Explanation:
To find the dimensions of the poster with the smallest area, we need to subtract the margins from the total dimensions of the poster and then find the dimensions that result in the smallest area. Let's assume the width of the poster is x cm and the height of the poster is y cm.
Using the information given, we can set up the following equations:
x - 2(6) = x - 12 cm (effective width)
y - 2(8) = y - 16 cm (effective height)
The area of the printed material is fixed at 390 square centimeters, so we have:
(x - 12) × (y - 16) = 390
To find the dimensions with the smallest area, we can find the derivative of the area equation with respect to either x or y, set it equal to zero, and solve for x or y. However, this is a complicated process. So, we can use a graphing calculator to find the minimum area. By substituting different values for x and y into the area equation, we can find the dimensions that result in the smallest area.
After substituting different values, we find that the dimensions of the poster with the smallest area are approximately 30 cm by 56 cm.
Julian walked 6/10 of a mile to his friends house and another 35/100 mile to the store. He walked 1/4 of a mile back home. Julian's sister said he walked 1/5 mile. Do you agreed. Why or why not
The statement, "Julian's sister said he walked 1/5 mile" cannot be agreed because Julian totally walked [tex]1\frac{1}{5} \text{ or } \frac{6}{5}[/tex] miles.
Solution:
Given that,
Julian walked 6/10 of a mile to his friends houseAnother 35/100 mile to the storeHe walked 1/4 of a mile back homeTo find total distance walked by Julian we have to add the above stated values. That is, [tex]\frac{6}{10} +\frac{35}{100} +\frac{1}{4}[/tex]
Factors of 10 = [tex]5\times2[/tex]
Factors of 100 = [tex]5\times2\times5\times2[/tex]
Factors of 4 = [tex]2\times2[/tex]
Therefore, the least common factor of 10, 100 and 4 is 100. With like denominators we can operate on just the numerators,
[tex]\frac{6\times10}{10\times10} +\frac{35\times1}{100\times1} +\frac{1\times25}{4\times1}\rightarrow\frac{60+35+25}{100}\rightarrow\frac{120}{100}[/tex]
[tex]\Rightarrow\frac{120}{100}\rightarrow\frac{6}{5}[/tex]
Which can also be written as [tex]1\frac{1}{5}[/tex].
So, from the above calculation it can be said that Julian walked [tex]1\frac{1}{5} \text{ miles }[/tex].
Julian did not walk 1/5 mile. He actually walked 1.2 miles.
Explanation:To determine whether Julian's sister's claim is accurate, we need to add up the distances Julian walked. He walked 6/10 mile to his friend's house, 35/100 mile to the store, and 1/4 mile back home. Using a common denominator of 100, we can add the fractions: 6/10 + 35/100 + 25/100 = 60/100 + 35/100 + 25/100 = 120/100 = 1.2 miles. Therefore, Julian walked 1.2 miles, not 1/5 mile as his sister claimed. So, I do not agree with his sister's statement.
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The probability of drawing two aces from a standard deck is 0.0059. We know this probability, but we don't know if the first card was replaced. If the two draws are defined as event A and event B, are the events dependent or independent?
A. They are dependent because, based on the probability, the first ace was replaced before drawing the second ace.
B. They are dependent because, based on the probability, the first ace was not replaced before drawing the second ace.
C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.
D. They are independent because, based on the probability, the first ace was not replaced before drawing the second ace.
Answer:
Option C is right
C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.
Step-by-step explanation:
Given that the probability of drawing two aces from a standard deck is 0.0059
If first card is drawn and replaced then this probability would change. By making draws with replacement we make each event independent of the other
Drawing ace in I draw has probability equal to 4/52, when we replace the I card again drawing age has probability equal to same 4/52
So if the two draws are defined as event A and event B, the events are independent
C. They are independent because, based on the probability, the first ace was replaced before drawing the second ace.
Find the area. The figure is not drawn to scale.
Answer:
1188 in²
Step-by-step explanation:
The area of a parallelogram is the product of its base length and height.
A = bh = (36 in)(33 in) = 1188 in²
Find the number of elements in A1 ∪ A2 ∪ A3 if there are 100 elements in A1, 1000 in A2, and 10,000 in A3 if
a) A1 ⊆ A2 and A2 ⊆ A3.
b) the sets are pairwise disjoint.
c) there are two elements common to each pair of sets and one element in all three sets.
(a) 1000
(b) 11100
(c) 11095.
Step-by-step explanation:
(a) If A1 is a subset of A2 and A2 is a subset of A3, then all the elements of A1 are in A2 and all the elements of A2 are in A3.
Then, n(A1 n A2) = 100, n(A2 n A3) = 1000 , n(A1 n A3) = 100 and n(A1 n A2 n A3) = 100.
So, we get
[tex]n(A1\cup A2\cup A3)\\\\=n(A1)+n(A2)+n(A3)-n(A1\cap A2)-n(A2\cap A3)-n(A1\cap A3)+n(A1\cap A2\cap A3)\\\\=100+1000+1000-100-1000-100+100\\\\=1000.[/tex]
(b) If the sets are pairwise disjoint, then
n(A1 n A2) = n(A2 n A3) = n(A1 n A3) = n(A1 n A2 n A3) = 0.
So, we get
[tex]n(A1\cup A2\cup A3)\\\\=n(A1)+n(A2)+n(A3)\\\\=100+1000+10000\\\\=11100.[/tex]
(c) If there are two elements common to each pair of sets and one element in all three sets, then
n(A1 n A2) = 2, n(A2 n A3) = 2, n(A1 n A3) = 2 and n(A1 n A2 n A3) = 1.
So, we get
[tex]n(A1\cup A2\cup A3)\\\\=n(A1)+n(A2)+n(A3)-n(A1\cap A2)-n(A2\cap A3)-n(A1\cap A3)-n(A1\cap A2\cap A3)\\\\=100+1000+1000-2-2-2+1\\\\=11100-5\\\\=11095.[/tex]
Final answer:
The number of elements in the union of sets A1, A2, and A3 varies depending on their relationships. For subsets (a), the count is 10,000; for disjoint sets (b), it is 11,100; and when each pair has common elements plus one common to all (c), the count is 11,095.
Explanation:
Finding the Number of Elements in the Union of Sets
To find the number of elements in the union of sets A1, A2, and A3, we need to consider the given conditions.
a) A1 ⊆ A2 and A2 ⊆ A3
Since A1 is a subset of A2, and A2 is a subset of A3, all elements of A1 and A2 are included in A3. Therefore, the
number of elements in A1 ∪ A2 ∪ A3 equals the number of elements in A3, which is 10,000.
b) The Sets Are Pairwise Disjoint
If the sets are pairwise disjoint, this means they share no elements in common. We simply add the number of elements in each set to find the union's total count. This gives us 100 + 1000 + 10,000 = 11,100 elements in the union.
c) Two Elements Common to Each Pair and One in All Three
With two elements common to each pair of sets and one element in all three, we need to subtract the common elements to avoid double-counting. So, A1 ∪ A2 ∪ A3 will have 100 + 1000 + 10,000 - 2 - 2 - 2 + 1 (since 1 element is counted three times, we add it back once) which equals 11,095 elements.
A line is parameterized by x=2+6???? and y=4+3????. (a) Which of the following points are on the section of the line obtained by restricting ???? to nonnegative numbers (for each, enter Y if the point is on the section, and N if not)? (−28,−11) : (8,7) : (26,16) : Then, give one more point that is on the section of the line obtained by this restriction: (b) What are the endpoints of the line segment obtained by restricting ???? to −2≤????≤1? left endpoint : right endpoint : (c) How should ???? be restricted to give the part of the line above the x-axis (give your answer as an interval for ????, for example, (3,8) or [-2,Inf))? ???? must be in :
Answer:
No, yes, yes
(-28,-11) and (8.7)
[tex][tex][\frac{-4}{3} ,\infty)[/tex]}[/tex]
Step-by-step explanation:
Given that a line in two dimension is parametrized by
[tex]x=2+6t \\y = 4+3t[/tex]
a) If t is non negative, then (-28,-11) cannot lie on that part
(-28,-11) No because t =-5
(8,7) yes because t =1
(26,16) yes because t = 4
b) when t lies between -2 and 1
we have left end point as
[tex]x=2+6(-2) = -10\\y = 4+3(-2) = -2\\[/tex]
(-10,-2) is left end point
Right end point is when t =1 i.e.
(8,7)
c) when the points should be above x axis, y should be non negative
i.e. [tex]y=4+3t\geq 0\\t\geq [/tex]
So t should lie in the interval
[tex][\frac{-4}{3} ,\infty)[/tex]}
Which number completes the inequality?
2/3 < <7/9
3/5
6/9
3/4
6/7
Answer:
3/4
Step-by-step explanation:
It can be helpful to use a common denominator for comparison. That denominator can be 100, meaning we can make them all decimal fractions. Approximate (2 digit) values are good enough for the purpose.
2/3 ≈ 0.67 . . . . left end of the range
7/9 ≈ 0.78 . . . . right end of the range
3/5 = 0.60
6/9 ≈ 0.67 . . . . = 2/3, so is not greater than 2/3
3/4 = 0.75
6/7 ≈ 0.86
The only decimal value between 0.67 and 0.78 is 0.75, corresponding to the fraction 3/4.
2/3 < 3/4 < 7/9
Let x represent one number and let y represent the other number. The sum of two numbers is negative 6. If one number is subtracted from the other, their difference is 8. Use the given conditions to write a system of equations. Solve the system and find the numbers.
Answer:
x=7 and y=-1
Step-by-step explanation:
X+Y=6 OR X=6-Y ...(1)
X-Y=8 ...(2)
substitue X=6-Y in (2)
(6-Y)-Y=8
6-2Y=8
-2Y=8-6
-2Y=2
Y=2/-2\Y=-1 ANS.
for x, substitute Y=-1 in (1) above
X-(-1)=8
X=8-1
X=7 ANS.
99 POINTS WILL GIVE BRAINLIEST!! No fake answers!
A computer programmer has a 35% chance of finding a bug in any given program. What is the probability that she finds a bug within the first three programs she examines?
A) 0.15
B) 0.27
C) 0.59
d) 0.73
A fair coin is flipped multiples times until it lands on heads. If the probability of landing on heads is 50%, what is the probability of first landing on heads on the fourth attempts?
A) 0.625
B) 0.0625
C) 0.500
D) 0.382
1.
Chance of finding a bug = 0.35
Chance of not finding a bug = 1 - 0.35 = 0.65
Probability of finding a bug in the first 3 programs =
Probability of not finding a bug in 2 out of the 3 and finding a bug in 1.:
0.65^2 * 0.35 = 0.147 = 0.15
Answer is A.
2.
Probability of heads = 0.50
Probability of tails = 0.50
Probability of heads on the fourth attempt = tails x tails x tails x heads = 0.5 x 0.5 x 0.5 x 0.5 = 0.0625
The answer is B.
Suppose that in a bushel of 100 apples there are 20 that have worms in them and 15 that have bruises. Only those apples with neither worms nor bruises can be sold. If there are 10 bruised apples that have worms in them, how many of the 100 apples can be sold?
To determine the number of sellable apples, we subtract the number of apples with defects from the total, but add back the ones counted twice due to having multiple defects. The calculation reveals that 75 out of 100 apples can be sold.
To calculate the number of apples that can be sold from the bushel, we need to consider those without worms or bruises. We have 20 apples with worms and 15 with bruises. However, since there are 10 apples that have both worms and bruises, these are counted twice in our total of defective apples.
First, we'll subtract the number of apples with worms (20) and those with bruises (15) from the total number of apples (100), but then we need to add back the ones we subtracted twice, those with both worms and bruises (10). Here's the calculation:
Total apples = 100
Apples with worms = 20
Apples with bruises = 15
Apples with both worms and bruises = 10
Apples that can be sold = Total apples - (Apples with worms + Apples with bruises - Apples with both worms and bruises)
Apples that can be sold = 100 - (20 + 15 - 10) = 100 - 25 = 75 apples can be sold.
75 of the 100 apples can be sold.
To find out how many apples can be sold, we need to determine the number of apples that are neither bruised nor have worms.
Given:
- Total number of apples = 100
- Number of apples with worms = 20
- Number of apples with bruises = 15
- Number of bruised apples with worms = 10
First, let's find the number of apples that have both bruises and worms. We are given that there are 10 bruised apples that have worms, so these apples are counted in both the bruised and worms categories. Therefore, we need to subtract these from the total number of bruised apples to avoid double-counting:
[tex]\[ \text{Number of apples with both bruises and worms} = 10 \][/tex]
Next, let's find the number of apples that have either bruises or worms. This can be done by adding the number of apples with bruises and the number of apples with worms and then subtracting the number of apples with both bruises and worms:
[tex]\[ \text{Number of apples with either bruises or worms} = 15 + 20 - 10 = 25 \][/tex]
Now, to find the number of apples that can be sold (i.e., the number of apples that are neither bruised nor have worms), we subtract the number of apples with either bruises or worms from the total number of apples:
[tex]\[ \text{Number of apples that can be sold} = 100 - 25 = 75 \][/tex]
So, 75 of the 100 apples can be sold.
A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. He needs no fencing along the river. What are the dimensions of the field that has the largest area?
Answer:
600ft x 1200ft
Step-by-step explanation:
Use derivative optimization to find the maximum area.
I'll call the two same sides "a", and the one different side "b"
The maximum perimeter (including 3 sides) is 2400 ft. so,
2400 = 2a + b
The area is length × width. so,
A = ab
Using substitution to combine the equations,
A = a × (2400 - 2a)
A = -2a² + 2400a
Find the maximum of A by finding the zeros of its derivative.
dA = -4a +2400
0 = -4a + 2400
The maximum occurs at a = 600
Substitute in the perimeter equation to find b.
2400 = 2(600) + b
b = 1200
600 x 1200
In a recent year, 32% of all college students were enrolled part-time. If 8.2 million college students were enrolled part-time that year, what was the total number of college students? Round your answer to the nearest million
Answer:
26 million
Step-by-step explanation:
8200000 / 0.32 = 25625000
25625000 rounded to nearest million = 26 million
The total number of college students is 26 million.
Given that, in a recent year, 32% of all college students were enrolled part-time and 8.2 million college students were enrolled part-time that year.
What is an equation?A mathematical equation is a formula that uses the equals sign to represent the equality of two expressions.
Let the total number of college students be x.
Now, 32% of x=8.2 million
⇒ 0.32 x=8200000
⇒ x = 8200000/3.2
⇒ x = 2562500
2562500 rounded to nearest million = 26 million
Therefore, the total number of college students is 26 million.
To learn more about an equation visit:
https://brainly.com/question/14686792.
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There are 9 showings of a film about endangered species at the science museum. A total of 459 people saw the film. The same number of people were at each showing. About how many people were at each showing?A. 40 and 50B. 50 and 60 C. 60 and 70 D. 70 and 80
Answer: B. 50 and 60
Step-by-step explanation:
Given : There are 9 showings of a film about endangered species at the science museum.
The total number of people saw the film = 459
Also, The same number of people were at each showing.
Then, the number of people were at each showing = Total people divided by Total showings
= 459 ÷ 9 = 51
Also, 50< 51 < 60 [the quotient is between 50 and 60.]
i.e. About 50 and 60 people were at each showing .
Hence, the correct answer is B. 50 and 60.