Suppose you're heading off for a long weekend (Friday, Saturday, and Sunday) somewhere and the weather report for your destination says: Chance of rain on Friday: 10% Chance of rain on Saturday: 25% Chance of rain on Sunday: 30% In each part below, find the chance exactly if it can be found using no further assumptions. If it can't be found, then (again using no further assumptions) find the best lower bound and upper bound that you can. a) the chance that it rains in your destination sometime during the long weekend

Answer:

The angle is 28.

Step-by-step explanation:

If we consider the system below (were "A"is the angle):  

A   +  [(x/2)-14] = 90       (complement)

A   +   x   = 180                 (suplement)

We can isolate "X=180 - A"  and apply this in the first equation. This way we are able to have only one variable, and w'll see that A=28.

Answer:

Her adjusted body weight is 78.26kg = 78.3kg, rounded to the nearest tenth.

Step-by-step explanation:

To find the Adjusted Body Weight(AjBW), first we have to consider the Ideal Body Weight(IBW).

We have the following formulas:

[tex]IBW = 45.5kg + 2.3kg*i[/tex], in which i is the number of inches that the woman has above 5 feet.

[tex]AjBW = IBW + 0.4*(ABW - IBW)[/tex], in which [tex]ABW[/tex] is her Actual Body Weight

So, in this problem:

Each pound has 0.45kg. So her weight is [tex]ABW = 220*0.45 = 99.8[/tex]kg.

The woman is 8 inches above 5 feet, so [tex]i = 8[/tex].

[tex]IBW = 45.5 + 2.3*8[/tex]

[tex]IBW = 63.9[/tex]kg

Her ideal body weight is 63.9kg. Her adjusted body weight is:

[tex]AjBW = IBW + 0.4*(ABW - IBW)[/tex]

[tex]AjBW = 63.9 + 0.4*(99.8-63.9)[/tex]

[tex]AjBw = 78.26[/tex]kg

Her adjusted body weight is 78.26kg = 78.3kg.

Answer:  60

Step-by-step explanation:

5 shorts × 4 shorts × 3 socks = 60 different uniforms

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

[tex]x_1\ =\ -2[/tex]

[tex]y_1\ =\ 3[/tex]

[tex]z_1\ =\ -4[/tex]

Distance is measure across the line

[tex]\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}[/tex]

So, we can write

[tex]\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k[/tex]

[tex]=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k[/tex]

[tex]=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k[/tex]

[tex]=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}[/tex]

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

[tex]3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3[/tex]

[tex]=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18[/tex]

[tex]=>18k-24+12k+9+10k-32\ =\ 18[/tex]

[tex]=>\ k\ =\dfrac{13}{8}[/tex]

So,

[tex]x\ =\ 3k-4[/tex]

   [tex]=\ 3\times \dfrac{13}{8}-4[/tex]

   [tex]=\ \dfrac{7}{4}[/tex]

[tex]y\ =\ \dfrac{4k+3}{2}[/tex]

   [tex]=\ \dfrac{4\times \dfrac{13}{8}+3}{2}[/tex]

   [tex]=\ \dfrac{19}{4}[/tex]

[tex]z\ =\ \dfrac{5k-16}{3}[/tex]

  [tex]=\ \dfrac{5\times \dfrac{13}{8}-16}{3}[/tex]

   [tex]=\ \dfrac{-21}{8}[/tex]

Now, the distance of point from the plane is given by,

[tex]d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}[/tex]

   [tex]=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}[/tex]

   [tex]=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}[/tex]

   [tex]=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}[/tex]

   [tex]=\ \sqrt{\dfrac{1177}{64}}[/tex]

   [tex]=\ 4.28[/tex]

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

            = 8.56 unit

So, the distance of a point from it's image is 8.56 units.

Answer:

The second one:

L∪J={i,j,k,l,m,n,o}

Step-by-step explanation:

The union is the elements listed in either set.

So since l,m,n, and o are elements of set L, they will also be elements of whatever it is "unioned" with.

Since i,j,k,l and m are elements of set J, they will also be elements of whatever it is "unioned" with.

When you write the union, just be sure to include each element that occurs in either set once.

So the union of L and J is {i,j,k,l,m,n,o}.

The answer is the second one.

The intersection would actually be that upside down U thing, the ∩ symbol. The intersection of two sets is a list of elements that both sets include.  So here the intersection would just consist of the elements l amd m.

The union of L and J is {i,j,k,l,m,n,o}. The correct option is the second option - L∪J = {i,j,k,l,m,n,o}

From the question, we are to determine the union of the sets

The union of two given sets is the smallest set which contains all the elements of both sets.

The given sets are

L= {l,m,n,o}

J= {i,j,k,l,m}

We are to determine the union of L and J, that is, L∪J

L∪J will be the set that contains all the elements of both L and J

Therefore

L∪J = {i,j,k,l,m,n,o}

Hence, the union of L and J is {i,j,k,l,m,n,o}. The correct option is the second option - L∪J = {i,j,k,l,m,n,o}

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Answer:

Step-by-step explanation:

First, observe that:

[tex](n+1)^2(2(n+1)^2-1)=(n^2+2n+1)(2n^2+4n+1)=2n^4+8n^3+11n^2+6n+1[/tex]

We will prove by mathematical induction that, for every natural,  

[tex]1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)[/tex]

We will prove our base case (when n=1) to be true.

Base case:

[tex]1^3+3^3+5^3+......(2n-1)^3=1=n^2(2n^2-1)[/tex]

Inductive hypothesis:  

Given a natural n,  

[tex]1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)[/tex]

Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

[tex]1^3+3^3+5^3+......(2(n+1)-1)^3=1^3+3^3+5^3+......(2n+1)^3=\\=n^2(2n^2-1)+(2n+1)^3=2n^4-n^2+8n^3+12n^2+6n+1=2n^4+8n^3+11n^2+6n+1[/tex]

Then, by the observation made at the beginning of this proof, we have that

[tex]1^3+3^3+5^3+......(2(n+1)-1)^3=(n+1)^2(2(n+1)^2-1[/tex]

With this we have proved our statement to be true for n+1.  

In conlusion, for every natural n

,

[tex]1^3+3^3+5^3+......(2n-1)^3=n^2(2n^2-1)[/tex]

b)

Observe that [tex]3\mid (n^3-n+3) \iff 3\mid (n^3-n) \iff 3\mid n(n^2-1)[/tex]

Then,

Therefore, for every [tex]n\in \mathbb{N}), 3\mid (n^3-n+3)[/tex]  

c)

We will prove by mathematical induction that, for every natural n>3,

[tex](n-1)^2<2n^2.[tex]

We will prove our base case (when n=4) to be true.

Base case:

[tex](n-1)^2=(4-1)^2=9<32=2*4^2=2n^2 [/tex]

Inductive hypothesis:  

Given a natural n>4,  

[tex](n-1)^2<2n^2.[tex]

Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.

Inductive step:

[tex]((n+1)-1)^2=((n-1)+1)^2=(n-1)^2+2(n-1)+1<2n^2+2(n-1)+1=2n^2+2n- 1<2n^2+2n+1 =2(n+1)^2.[tex]

With this we have proved our statement to be true for n+1.    

In conclusion, for every natural n>3,

[tex](n-1)^2<2n^2.[tex]

Answer: 1/15 of all the workers invest the maximum allowable to the retirement plan.

Step-by-step explanation:

x = hourly workers

y = salaried workers

1/6 of the hourly workers invest the  retirement plan  = 1/6 * x = x/6

1/2 of the salaried workers invest in the same  plan. = 1/2 * y = y/2

1/5 of the hourly workers invest the maximum allowed = 1/5 * x/6 = x/30

1/3 of the  salaried workers invest the maximum allowed = 1/3 * y/2 = y/6

there are three times as many hourly workers as salaried  workers

x = 3y

Total workers: x + y = 3y + y = 4y

All workers that invest the maximum allowed: x/30 + y/6 = 3y/30 + y/6 =

3y + 5y = 8y = 4y

   30       30    15  

If all the workers are 4y, then 1/15 of all the workers invest the maximum allowable to the retirement plan.

Answer:

Step-by-step explanation:

You already have the answer.

1 [at] = 760 [mmHg]

1 [at] = 14,7 [psi]

1500 [mmHg] * 1 [at] / 760 [mmHg] * 14,7 [psi] / 1 [at] = 29.0 [psi]

recall your d = rt, distance = rate * time.

A = first plane

B = second plane

the assumption is that they both start off from the same point at the same time in opposite directions.

at some time say "t" hours, plane A will be some distance, hmmm say "d" miles, since we know they'll both be 4000 miles apart, then plane B will be "4000 - d" miles at that time, since both took off at the same time, then plane A as well as plane B have both been travelling "t" hours.

[tex]\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{plane A}&d&400&t\\ \textit{plane B}&4000-d&600&t \end{array}\qquad \qquad \begin{cases} \boxed{d}=400t\\\\ 4000-d=600t \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting on the 2nd equation}}{4000-\boxed{400t}=600t}\implies 4000=1000t\implies \cfrac{4000}{1000}=t\implies 4=t[/tex]

Final answer:

There are 21 different sets of five marbles that can be made from one transparent, three yellow, and three orange marbles, with no red or green marbles included.

Explanation:

The student has asked how many possible sets of five marbles there are, with the restriction that none of the marbles in a set can be red or green. Considering the available marbles, the student can only choose from one transparent, three yellow, and three orange marbles.

Since there is no replacement and the colors do not matter beyond not being red or green, the student is creating combinations of seven unique marbles taken five at a time. This can be calculated using the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of items to choose from, k is the number needed for the set, and ! denotes factorial.

For this problem, n is 7 (1 transparent + 3 yellow + 3 orange) and k is 5. So, we calculate C(7, 5) = 7! / (5!(7 - 5)!) = 7! / (5!2!) = (7 × 6) / (2 × 1) = 21. There are 21 different sets of five marbles where none are red or green.

Answer:

[tex]a.\hspace{3} \frac{6}{38} = 0.15789474\\\\b.\hspace{3} \frac{45}{55} = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 0.14285714\\\\d.\hspace{3} \frac{35}{28} = 1.25\\\\e.\hspace{3} \frac{1030}{2030} = 0.50738916[/tex]

Step-by-step explanation:

Rational numbers, expressed as decimal numbers, are obtained from the operation of division between the integer of the numerator and the integer of the denominator. Then:

[tex]a.\hspace{3} \frac{6}{38} = 6\div38 = 0.15789474\\\\ b.\hspace{3} \frac{45}{55} = 45\div55 = 0.81818181\\\\c.\hspace{3} \frac{4}{28} = 4\div28 = 0.14285714\\\\ d.\hspace{3} \frac{35}{28} = 35\div28 = 1.25\\\\ e.\hspace{3} \frac{1030}{2030} = 1030\div2030 = 0.50738916[/tex]

Answer:

7438.

Step-by-step explanation:

Let x be the attendance before the drop.

We have been attendance dropped 16% this year, to 6248. We are asked to find the attendance before the drop.

The attendance after drop would be 84% (100%-16%) of x.

[tex]\frac{84}{100}\cdot x=6248[/tex]

[tex]0.84x=6248[/tex]

[tex]\frac{0.84x}{0.84}=\frac{6248}{0.84}[/tex]

[tex]x=7438.095[/tex]

[tex]x\approx 7438[/tex]

Therefore, the attendance before the drop is 7438.

Answer:

See explanation

Step-by-step explanation:

During daytime in New York City, UberRUSH has a base fare of $3.00 plus $2.50 per mile.

A. Let x be the number of miles Sergio uses the taxi, then he will pay $2.50x for x miles plus a base fare of $3.00. In total, the cost of using UberRUSH services during the day in New York City is

[tex]C=3.00+2.50x[/tex]

B. The rate of change for the cost of riding UberRUSH during the day in New York City is the slope of the line represented by linear function in part A. So, the rate of change is 2.5 and it shows the change in price per 1 mile driven.

C. The y-intercept is 3.00 (or simply 3) and it represents the initial cost (when 0 miles are driven)

D. The graph of the function is shown in attached diagram. The diagram shows the part of the line starting from point (0,3). This is because the number of miles driven cannot be negative.

Answer:

The sum of even numbers between 1 and 100 is 2550.

Step-by-step explanation:

To find : Add the even numbers between 1 and 100?

Solution :

The even numbers from 1 to 100 is 2,4,6,...,100  form an arithmetic progression,

The first term is a=2

The common difference is d=2

The last term is l=100

First we find the number of terms given by,

[tex]l=a+(n-1)d[/tex]

[tex]100=2+(n-1)2[/tex]

[tex]100=2+2n-2[/tex]

[tex]2n=100[/tex]

[tex]n=\frac{100}{2}[/tex]

[tex]n=50[/tex]

The sum formula of A.P is

[tex]S_n=\frac{n}{2}[a+l][/tex]

Substitute the values in the formula,

[tex]S_{50}=\frac{50}{2}[2+100][/tex]

[tex]S_{50}=25\times 102[/tex]

[tex]S_{50}=2550[/tex]

Therefore, The sum of even numbers between 1 and 100 is 2550.

Answer:

8

Step-by-step explanation:

Maximum amount of Acetaminophen that can be taken = 4 g per day

Weight of acetaminophen tablet = 500 mg

let the number of tablets that can be taken be 'x'

therefore,

x × 500 mg ≤ 4 g

also, 1 g = 1000 mg

thus,

x × 500 ≤ 4000

or

x ≤  8

hence,

the maximum numbers of tablets that can be taken per day is 8

Answer:

a)109 of the surveyed households only own a hand gun and do not support the ban on hand guns.

b)328 of the surveyed households do not own a gun and support the ban on hand guns

c )112 of the surveyed households do not own a gun and do not support the ban on handguns.

Step-by-step explanation:

To solve this problem, we must build the Venn's Diagram of this set.

I am going to say that:

-The set A represents those who own a hand gun.

-The set B represents those who own a rifle

-The set C represents those who support the ban of handguns.

-D represents those who do not own a gun and do not support the ban on hand guns.

We have that:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

In which [tex]a[/tex] are those that only own a hand gun, [tex]A \cap B[/tex] are those who own both a handgun and a rifle, [tex]A \cap C[/tex] are those who own a hand gun and support the ban on handguns, and [tex]A \cap B \cap C[/tex] are those who own both a hand gun and a rifle, and also support the ban on hand guns.

By the same logic, we have:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

In which [tex]b[/tex] are those who only own a rifle, and [tex]B \cap C[/tex] are those who own a rifle and support the ban on handguns.

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

In which [tex]c[/tex] are those who support the ban on hand guns and do not own any handguns.

This diagram has the following subsets:

[tex]a,b,c,D(A \cap B), (A \cap C), (B \cap C), (A \cap B \cap C)[/tex]

There were 1000 households surveyed, so:

[tex]a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1000[/tex]

We start finding the values from the intersection of three sets.

Solution:

48 own both a hand gun and a rifle and also support the ban on hand guns.

[tex]A \cap B \cap C = 48[/tex]

70 own a hand gun and support the ban on hand guns:

[tex](A \cap C) + (A \cap B \cap C) = 70[/tex]

[tex]A \cap C = 70 - 48[/tex]

[tex]A \cap C = 22[/tex]

206 own a rifle but no hand gun and do not support the ban on hand guns.

[tex]b = 206[/tex].

136 own both a hand gun and a rifle:

[tex](A \cap B) + (A \cap B \cap C) = 136[/tex]

[tex]A \cap B = 136 - 48[/tex]

[tex]A \cap B = 88[/tex]

493 supported the ban on hand guns.

[tex]C = 493[/tex]

437 own a rifle:

[tex]B = 437[/tex]

267 own a hand gun:

[tex]A = 267[/tex]

a) How many of the surveyed households only own a hand gun and do not support the ban on hand guns?

This is the value of [tex]a[/tex], that we can find from the following equation:

[tex]A = a + (A \cap B) + (A \cap C) + (A \cap B \cap C)[/tex]

[tex]267 = a + 88 + 22 + 48[/tex]

[tex]a = 267 - 158[/tex]

[tex]a = 109[/tex]

109 of the surveyed households only own a hand gun and do not support the ban on hand guns.

b) How many of the surveyed households do not own a gun and support the ban on hand guns?

This is the value of [tex]c[/tex], that we can find from the following equation:

[tex]C = c + (A \cap C) + (B \cap C) + (A \cap B \cap C)[/tex]

[tex]493 = c + 22 + (B \cap C) + 48[/tex]

Here, we also have to find [tex]B \cap C[/tex], that are those who own a rifle and support the ban on hand guns. We can find this from the following equation:

[tex]B = b + (B \cap C) + (A \cap B) + (A \cap B \cap C)[/tex]

[tex]437 = 206 + (B \cap C) + 88 + 48[/tex]

[tex]B \cap C = 437 - 342[/tex]

[tex]B \cap C = 95[/tex]

So

[tex]493 = c + 22 + (B \cap C) + 48[/tex]

[tex]493 = c + 22 + 95 + 48[/tex]

[tex]c = 493 - 165[/tex]

[tex]c = 328[/tex]

328 of the surveyed households do not own a gun and support the ban on hand guns.

a) How many of the surveyed households  do not own a gun and do not support the ban on hand guns?

This is the value of d, that can be found from the following equation:

[tex]a + b + c + D + (A \cap B) + (A \cap C) + (B \cap C) + (A \cap B \cap C) = 1000[/tex]

[tex]109 + 206 + 328 + D + 88 + 22 + 95 + 40 = 1000[/tex]

[tex]D = 1000 - 888[/tex]

[tex]D = 112[/tex]

112 of the surveyed households do not own a gun and do not support the ban on handguns.

I like guns but im christian

To measure tartar emetic for the prescription, keep in mind the sensitivity of your balance and the acceptable weighing error. Tartar emetic of 15mg should be weighed as accurately as the balance allows and if the measurement falls within the acceptable error range, it can be dissolved in cherry syrup for uniform distribution.

The tartar emetic dose required by the prescription is 0.015 g. However, you're using a balance with a sensitivity of 4 mg (milligrams). Similar to procure the accurate tartar emetic measurement, you must take into account the balance's sensitivity and the acceptable weighing error of 5%.

0.015 g is equal to 15 mg. Our balance's lowest measurement is 4 mg, which is less than the 15 mg we need to weight accurately. When considering the acceptable weighing error of 5%, the actual amount of tartar emetic measured could range from 14.25 mg to 15.75 mg (15 mg ± 5%).

To solve with the stipulated constraints, you need to carefully weigh out the tartar emetic as close as the balance will allow, and if it does not reach the desired measurement but falls within the acceptable error range, you can dissolve it in the cherry syrup for uniform distribution within the medication. Always remember to mix well and be precise with measurements to ensure the correct strength of medicine.

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Calculate the acceptable error range for the weight of tartar emetic, then weigh it out on a balance with the specified sensitivity. Once within the error range, dissolve the tartar emetic in the cherry syrup.

The question asks how to obtain the correct quantity of tartar emetic given the equipment and accepted weighing error. To do this, first you must calculate the acceptable error range by multiplying the desired weight of the tartar emetic (0.015 g) by 5%, which yields an acceptable error of 0.00075 g. Since your balance has a sensitivity of 4 mg (0.004 g), it can accurately measure out this quantity. Then you would weigh out the tartar emetic on the balance until you are within this error range.

It's important to note that tartar emetic is fully soluble in water and so cherry syrup, which is water-based, can be used as the solvent for tartar emetic. After weighing, dissolve the tartar emetic into the cherry syrup before moving on to measure out the other ingredients for the prescription.

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Answer:

The answer is 36 degrees

Step-by-step explanation:

Lets call the variables X= Base Angle and Y= Vertix Angle and  X=2Y

As Isosceles triangle theorem "If two sides of a triangle are congruent, then the angles opposite those sides are congruent" We asume that the X value of each base angle is the same and the sum of the three angles are equal to 180 degrees, so we have:

1)X+Y+X=180

2) We know that X=2Y so we replace them in the below formula

3) 2Y+Y+2Y=180

4) 5Y=180

5) Then we resolve the variable Y and divide the 180 degrees by 5 Y=180/5

6)Then we have that Y=36 Degrees

Please see attachment to follow up the step by step

I hope that this answer finds you well

The only way to get a term containing [tex]x^{25}[/tex] is to take 6 copies of 1, 1 copy of [tex]x[/tex], and 3 copies of [tex]x^8[/tex] (since 6+1+3=10, and 6*0+1*1+3*8=25). So this term has a coefficient of

[tex]\dfrac{10!}{6!1!3!}=840[/tex]

Answer:

a) There are 11 balls left in the bag.

b) 4 of the balls in the bag are red.

c) 3 brown, 2 green, 2 white.

d) [tex]P = \frac{2}{11}[/tex]

e) The probability that the first ball is red and the second is green is:

[tex]\frac{5}{66}[/tex]

The probability that both balls are red is:

[tex]\frac{5}{33}[/tex]

f) The probability that both balls are green is:

[tex]\frac{1}{66}[/tex]

g) The probability that the first ball is brown and the second is white is:

[tex]\frac{1}{22}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]\frac{1}{22}[/tex]

They are the same probabilities.

Step-by-step explanation:

There are 12 balls in the bag

5 red

3 brown

2 green

2 white

If you draw a red ball and put it aside:

a) How many balls are left in the bag?

There were 12 balls in the bag, and one was put aside.

So, there are 11 balls left in the bag.

b) How many of the balls in the bag are red?

There were 5 red balls in the bag, and one was put aside.

So, there are 4 red balls in the bag.

c) How many are brown? Green? White?

The ball put aside was red, so we still have the same number of the balls of the other colors.

3 brown, 2 green, 2 white.

d) Draw a second ball from the bag. What is the probability that it is green?

There are 11 balls in the bag, 2 of which are green. So the probability that the second ball is green is:

[tex]P = \frac{2}{11}[/tex]

Now put both balls back in the bag and draw two balls without replacing them.

e) What is the probability that the first ball is red and the second ball is green (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is red. There are 12 balls, 5 of which are red. So:

[tex]P_{1} = \frac{5}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 2 of which are green. So:

[tex]P_{2} = \frac{2}{11}[/tex]

The probability that the first ball is red and the second is green is:

[tex]P = P_{1}*P_{2} = \frac{5}{12}*\frac{2}{11} = \frac{5}{66}[/tex]

What is the probability that the first ball is red and the second ball is also red (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is red. There are 12 balls, 5 of which are red. So:

[tex]P_{1} = \frac{5}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 4 of which are red. So:

[tex]P_{2} = \frac{4}{11}[/tex]

The probability that both balls are red is:

[tex]P = P_{1}*P_{2} = \frac{5}{12}*\frac{4}{11} = \frac{5}{33}[/tex]

f) What is the probability that the first ball is green and the second ball is also green (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is green. There are 12 balls, 2 of which are green. So:

[tex]P_{1} = \frac{2}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 1 of which is green. So:

[tex]P_{2} = \frac{1}{11}[/tex]

The probability that both balls are green is:

[tex]P = P_{1}*P_{2} = \frac{2}{12}*\frac{1}{11} = \frac{1}{66}[/tex]

g) What is the probability that the first ball is brown and the second ball is white (without replacement)?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is brown. There are 12 balls, 3 of which are brown. So:

[tex]P_{1} = \frac{3}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 2 of which are white. So

[tex]P_{2} = \frac{2}{11}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]P = P_{1}*P_{2} = \frac{3}{12}*\frac{2}{11} = \frac{1}{22}[/tex]

Would this be different than the probability that the first ball is white and the second ball is brown?

[tex]P = P_{1}*P_{2}[/tex]

[tex]P_{1}[/tex] is the probability that the first ball is white. There are 12 balls, 2 of which are brown. So:

[tex]P_{1} = \frac{2}{12}[/tex]

Since there are no replacements, now there are 11 balls in the bag, 3 of which are brown. So

[tex]P_{2} = \frac{3}{11}[/tex]

The probability that the first ball is brown and the second is white is:

[tex]P = P_{1}*P_{2} = \frac{2}{12}*\frac{3}{11} = \frac{1}{22}[/tex]

They are the same probabilities.

Answer:

3.045%.

Step-by-step explanation:

We are asked to find the corresponding effective interest rate for 3% per year compounded continuously.

We will use effective interest formula to solve our given problem.

[tex]r=e^i-1[/tex], where,

r = Effective interest rate,

e = Mathematical constant,

r = Interest rate in decimal form.

Let us convert given interest rate in decimal form.

[tex]3\%=\frac{3}{100}=0.03[/tex]

Substitute values:

[tex]r=e^{0.03}-1[/tex]

[tex]r=1.030454533953517-1[/tex]

[tex]r=0.030454533953517[/tex]

[tex]r\approx 0.03045[/tex]

Convert into percentage:

[tex]0.03045\times 100\%=3.045\%[/tex]

Therefore, the corresponding interest rate would be 3.045%.

Answer:

$5250 net sales were generated by the sandals.

Step-by-step explanation:

Given :

The shoe department had yearly net sales of $375,000.

Sandals represented 1.4 % of the total net sales.

To Find : What net sales dollars were generated by the sandal ?

Solution:

The shoe department had yearly net sales of $375,000.

Sandals represented 1.4 % of the total net sales.

So,  net sales dollars were generated by the sandal = [tex]1.4 \% \times 375000[/tex]

Net sales dollars were generated by the sandal = [tex]\frac{1.4}{100} \times 375000[/tex]

                                                                              = [tex]5250[/tex]

Hence $5250 net sales were generated by the sandals.

Answer:

600 ml

Step-by-step explanation:

Given,

240 ml of milk provides about 8 grams of protein,

That is, the ratio of grams of protein provided and milk = [tex]\frac{8}{240}[/tex]

Let x be the quantity of milk in ml that provides 20 grams of protein,

So, the ratio of grams of protein provided and milk = [tex]\frac{20}{x}[/tex]

[tex]\implies \frac{8}{240}=\frac{20}{x}[/tex]

By cross multiplication,

8x = 4800

Divide both sides by 8,

x = [tex]\frac{4800}{8}[/tex] = 600

Hence, 600 ml milk should be needed to provide 20 grams of protein.

To obtain 20 grams of protein, you would need to get 600 ml of milk.

To find out how much milk you should get to obtain 20 grams of protein, we can use a proportion:

240 ml of milk provides 8 grams of protein

x ml of milk provides 20 grams of protein

To solve for x, we can set up the following proportion:

240/8 = x/20

Cross multiplying, we get:

8x = 240 * 20

x = (240 * 20)/8

Therefore, you would need to get 600 ml of milk to obtain 20 grams of protein.

Answer:

[tex]angle = 0.24 rad = 14 °[/tex]

Step-by-step explanation:

The angle of depression is the angle formed by the horizontal at the camera position and the line formed by the camera and the objective, therefore it can be calculated from the information provided as shown in the attached file:

First, you draw f' and f" as parallel lines to the floor at the height of the objective and the height of the camera respectively. Then draw the line o between camera and objective.

the blue angle created by o and f" is the depression angle, which is the same as the angle created by o and f' because angles between parallel lines.

You need to calculate b, as:

b = h - a = 9 - 6 = 3

Then, for the trigonometric function tangent as we have a rectangle triangle:

[tex] Tan(angle)=\frac{b}{f'}[/tex]

therefore:

[tex]angle = Tan^{-1}(\frac{b}{f'} ) = Tan^{-1}(\frac{3}{12} )[/tex]

[tex]angle = 0.24 rad = 14 °[/tex]

Answer:

There is approximately 5.475 liters of blood in the body.

Step-by-step explanation:

At a resting pulse rate of 75 beats per minute, the human heart typically pumps about 73 ml of blood per beat.

73 ml in liters = [tex]73\times0.001=0.073[/tex] liters

Now, 75 beats/min x  0.073 liters/beat  = 5.475 liters/min

And 5.475 liters/min x 1 min/body = 5.475 liters/body

Hence, there is approximately 5.475 liters of blood in the body.

Answer:  374416

Step-by-step explanation:

Given : A test requires that you answer first Part A and then either Part B or Part C.

Part A consists of 4 true false questions, Part B consists of 6 multiple-choice questions with one correct answer out of five, and Part C consists of 5 multiple-choice questions with one correct answer out of six.

i.e. 2 ways to answer each question in Part A.

For 4 questions, Number of ways to answer Part A = [tex]2^4[/tex]

5 ways to answer each question in Part B.

For 6 questions, Number of ways to answer Part B = [tex]5^6[/tex]

6 ways to answer each question in Part C.

For 5 questions, Number of ways to answer Part C = [tex]6^5[/tex]

Now, the number of ways to  completed answer sheets are possible :_

[tex]2^4\times5^6+2^4\times6^5\\\\=2^4(5^6+6^5)\\\\=16(15625+7776)\\\\=16(23401)=374416[/tex]

Hence, the number of ways to  completed answer sheets are possible = 374416

Answer:

325/790 rounded to the nearest hundredths place is 0.41

Step-by-step explanation:

The given fraction is :

[tex]\frac{325}{790}[/tex]

Dividing by 5:

[tex]\frac{65}{158}[/tex]

= 0.41139

We can see that we have a 1 after the hundredth place, so we will not round the 41 as 42.

Now, rounding this to the nearest hundredths place, we get 0.41.

Answer:

[tex]x=\frac{\pi}{4}[/tex] and [tex]x=-\frac{3\pi}{4}[/tex]

Step-by-step explanation:

We are given that sin x+cos x=cos 2x

We have to solve the given equation for [tex]0\leq x\leq 2\pi[/tex]

[tex] sin x+cos x=cos^2x-sin^2x[/tex]

Because [tex]cos 2x=cos^2-sin^2[/tex]

[tex] sinx+cos x=(sinx +cos x)(sinx-cos x)[/tex]

[tex] 1 =sin x-cos x[/tex]

[tex] sin x=cos x[/tex]

[tex]\frac{sinx }{cos x}=1[/tex]

[tex]tan x=1[/tex]

[tex]tan x=\frac{sinx}{cos x}[/tex]

[tex]tan x=tan\frac{\pi}{4}[/tex]

[tex]x=\frac{\pi}{4}[/tex]

Tan x is positive in I and III quadrant

In III quadrant angle[tex]\theta[/tex]  replace by [tex]\theta -\pi[/tex]

Therefore, [tex]tan x=tan (\frac{\pi}{4}-\pi)=tan\frac{\pi-4\pi}{4}=tan\frac{-3\pi}{4}[/tex]

[tex]x=-\frac{3\pi}{4}[/tex]

Answer:

A) find its center  = (3, 2)

B) find the radius of the circle = √104

C) find the equation of the circle = x² + y² - 6x - 4y -91 = 0

Step-by-step explanation:

A)- The center must be the mid-points of (-2, 1) and (8, 3).

So, using the equation of mid-point,

[tex]h=\frac{x_{1}+x_{2}}{2}  and  k=\frac{y_{1}+y_{2}}{2}[/tex]

Here, (x₁, y₁) = (-2, 1) and (x₂, y₂) = (8, 3)

Putting these value in above equation. We get,

h = 3 and k = 2

Thus, Center = (h, k) = (3, 2)

B)- For finding the radius we have to find the distance between center and any of the end point.

Thus using Distance Formula,

[tex]Distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]

[tex]Radius =\sqrt{(8+2)^{2}+(3-1)^{2}}[/tex]

⇒ Radius = √104 = 2√26

C)- The equation of circle is determined by formula:

[tex](x-h)^{2}+(y - k)^{2} = r^{2}[/tex]

where (h, k ) is center of circle and

r is the radius of circle.

⇒ (x - 3)² + (y - 2)² = 104

⇒ x² + y² - 6x - 4y -91 = 0

which is the required equation of the circle.

The mean cost of repair is $391.75. The median cost of repair is $419.50. There is no mode for the repair costs.

To compute the mean, median, and mode cost of repair, we first need to list the repair costs in numerical order:

The mean is calculated by finding the average of the repair costs, which is the sum of the costs divided by the total number of costs. In this case, the mean is $(249 + 403 + 436 + 479)/4 = $391.75.

The median is the middle value of the data set once it is ordered. In this case, the median is $419.50, which is the average of $403 and $436.

The mode is the value that appears most frequently in the data set. In this case, there is no mode because none of the repair costs occur more than once.

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The mean cost of repair is $391.75, while the median cost is $419.50. Since there are no repeated repair costs, there is no mode.

To determine the mean, median, and mode repair costs of the four cars, follow these steps:

Mean

The mean (average) cost of repair is calculated by summing all the repair costs and then dividing by the number of cars.

Sum of costs = $436 + $403 + $479 + $249 = $1567

Number of cars = 4

Mean = $1567 / 4 = $391.75

So, the mean cost of repair is $391.75.

Median

To find the median, list the repair costs in ascending order: $249, $403, $436, $479.

Since there are four data points (even number), the median is the average of the two middle numbers.

Median = ($403 + $436) / 2 = $839 / 2 = $419.50

So, the median cost of repair is $419.50.

Mode

The mode is the number that appears most frequently in a data set.

In this case, each cost appears only once; therefore, no mode exists.