Suppose that on the average, 7 students enrolled in a small liberal arts college have their automobiles stolen during the semester. What is the probability that less than 1 student will have his automobile stolen during the current semester

Answer:

To find the surface area of a cuboid we can also label the length, width and the height of the prism and use the formula SA=2LW+2LH+2HW to find the area of a cuboid

Answer:

202 cm²

Step-by-step explanation:

The opposite faces of a cuboid are congruent , then

SA = top/bottom + front/ back + sides , that is

SA = 2(9 × 4) + 2(9 × 5) + 2(4 × 5)

     = 2(36) + 2(45) + 2(20)

     = 72 + 90 + 40

    = 202 cm²

Answer:

In any quantitative science, the terms relative change and relative difference are used to compare two quantities while taking into account the "sizes" of the things being compared. The comparison is expressed as a ratio and is a unitless number.

Step-by-step explanation:

I hope it helps

Answer:

2 feet

Step-by-step explanation:

Displacement at 0 seconds is 0 feet.

Displacement at 2 seconds is 2 feet because it took them 2 seconds to walk 2 feet.

Answer:

x= - 1

Step-by-step explanation:

I think you are asked to find the value of the first derivative of f(x) at π/4. Given

[tex]f(x) = \dfrac{\sin(4x)-\cos(x)}{\tan(x)}[/tex]

use the quotient to differentiate and you get

[tex]f'(x) = \dfrac{\tan(x)(4\cos(4x)+\sin(x))-(\sin(4x)-\cos(x))\sec^2(x)}{\tan^2(x)}[/tex]

Then at x = π/4, you have

tan(π/4) = 1

cos(4•π/4) = cos(π) = -1

sin(π/4) = 1/√2

sin(4•π/4) = sin(π) = 0

cos(π/4) = 1/√2

sec(π/4) = √2

==>   f ' (π/4) = (1•(-4 + 1/√2) - (0 - 1/√2)•(√2)²) / 1² = -4 + 1/√2 + √2

Answer:

[tex]{ \tt{rate \: of \: change \: in \: A = 9}}[/tex]

Rate of change in function A is two times than that in function B

Answer:

22608 mm³/s

Step-by-step explanation:

Applying chain rule,

dV/dt = (dV/dr)(dr/dt)............... Equation 1

Where dV/dr = rate at which the volume is increasing

But,

V = 4πr³/3

Therefore,

dV/dr = 4πr²............... Equation 2

Substitute equation 2 into equation 1

dV/dt = 4πr²(dr/dt).............. Equation 3

From the question,

Given: dr/dt = 2 mm/s, r = 60/2 = 30 mm

Consatant: π = 3.14

Substitute these values into equation 3

dV/dt = 4×3.14×30²×2

dV/dt = 22608 mm³/s

Answer:

121 unit^2.

Step-by-step explanation:

The area = height/2 * ( sum of the opposite parallel  lines)

= h/2(BC + AD

h = BF = 14 - 3 = 11 units.

BC   = 13 - 5 = 8 units.

AD = 16 - 2 = 14 units.

Area = (11/2)(8 + 14)

= 5.5 * 22

= 121 unit^2.

Answer:

121

Step-by-step explanation:

Answer:

283 flowers

Step-by-step explanation:

c=2pi*r

c = 1130.973 =1131

1131/4

282.75 = 283

Answer:

[tex]\frac{y^{6} }{ x^{2} }[/tex]

Step-by-step explanation:

[tex]y^{6} x^{-2}[/tex]

Answer and Step-by-step explanation:

When there is a set of values within parenthesis, and an exponents on the values and on the parenthesis, you multiply the outer exponent with the inner exponent.

When a value has a negative exponent, the value that has the negative exponent will become a fraction and go to the denominator of the fraction (or go immediately to the denominator), or if it is already a fraction, goes to the denominator. If the value that has the negative exponent is in the denominator, the value will go to the numerator. In both instances, the negative exponent will then change to positive.

First, we need to simplify the expression inside the parenthesis.

[tex]y^{\frac{3}{2} } x^{-\frac{1}{2} } --> \frac{y^{\frac{3}{2} } }{x^{\frac{1}{2}} }[/tex]

Now we multiply the 4 to the exponents.

[tex]\frac{y^{\frac{3}{2} *\frac{x4}{1} } }{x^{\frac{1}{2}}*\frac{4}{1} } = \frac{y^{\frac{12}{2}} }{x^{\frac{4}{2}}} = \frac{y^6}{x^2}[/tex]

[tex]\frac{y^6}{x^2}[/tex] is the answer.

#teamtrees #PAW (Plant And Water)

Answer:

1/108

Step-by-step explanation:

each denominator triples, so just triple 36.

Answer:

1/108

Step-by-step explanation:

This is a geometric sequence, where each number is 3 times the previous. Normally you would use the actual formula, however you're just asked to pick up on a pattern so just multiplying the second number by 3 works.

Actual data table :

X __ y

2 15

6 13

7 9

8 8

12 5

Answer:

0.953

Step-by-step explanation:

The question isnt well formatted :

The actual data:

X __ y

2 15

6 13

7 9

8 8

12 5

Using a correlation Coefficient calculator, the correlation Coefficient obtained by fitting the data is 0.953 which depicts a strong linear correlation between the x and y variable. This shows that the value of y increases with a corresponding increase in x values and vice versa.

Answer:

Addition equation = -4-0) + [(-13)-(-4)]

Answer =  -13

Step-by-step explanation:

For the small arrow in the diagram, the expression is (-4 - 0)

For the bog arrow, the expression will be -13 - (-4)

Adding both expressions

Addition = (-4-0) + [(-13)-(-4)]

Addition = (-4) + (-13+4)

Addition = -4 + (-9)

Addition = -4-9

Addition = -13

Answer:

129469.3194

342000

212530.6806

Step-by-step explanation:

Going to assume that the 8% is a nominal, montly rate

which means the effective monthly rate is .08/12= .006667

using the annuity immediate formula...

a.)

[tex]950(\frac{1-(1+.006667)^{-30*12}}{.006667})=129469.3194[/tex]

b.) we would pay 950*30*12= 342000

c.) the amount in interest would be 342000-129469.3194=212530.6806

a) The loan one can afford is $1,29,460.2

b) The total amount of money paid to the loan company over the life of the loan is $342,000.

c)  $212539.8 of the total amount paid is interest.

To determine the answers to these questions, we'll need to use the formula for calculating a fixed monthly mortgage payment:

[tex]M = \frac{P \times r \times (1 + r)^n}{((1 + r)^n - 1)}[/tex]

where:

M is the monthly payment,

P is the principal loan amount,

r is the monthly interest rate (annual interest rate divided by 12),

and n is the total number of payments (number of years multiplied by 12).

Given:

Monthly payment (M) = $950

Loan term = 30 years

Interest rate = 8% per year

a) How big of a loan can you afford?

Let's calculate the principal loan amount (P):

First, we need to convert the annual interest rate to a monthly interest rate:

r = 0.08 / 12

= 0.00667

n = 30 years × 12 months

n= 360

Using the formula and plugging in the values we have:

[tex]950 = \frac{P \times 0.00667 \times (1 + 0.00667)^{360}}{((1 + 0.00667)^{360} - 1)}[/tex]

[tex]950 = \frac{P \times 0.00667 \times 10.948}{10.948 - 1}[/tex]

[tex]950=\frac{P \times 0.07302316}{9.948}[/tex]

[tex]950\times9.948 = 0.0730P[/tex]

Divide by 0.073:

Now we can solve for P:

[tex]P=\frac{9450.6}{0.0730}[/tex]

[tex]P = 1,29,460.2[/tex]

Therefore, you can afford a loan amount of $1,29,460.2

b) The total amount paid to the loan company can be calculated by multiplying the monthly payment by the total number of payments:

Total amount = Monthly payment × Total number of payments

Total amount =[tex]$950 \times 360[/tex]

Total amount = [tex]342,000[/tex]

Therefore, the total amount of money paid to the loan company over the life of the loan is $342,000.

c) To find out how much of the total amount paid is interest, we can subtract the principal loan amount from the total amount:

Interest = Total amount - Principal loan amount

Interest = [tex]342,000 - 129460.2[/tex]

=$212539.8

Therefore, $212539.8 of the total amount paid is interest.

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Answer:

[tex]m = \frac{1}{12}[/tex]

Step-by-step explanation:

Given

[tex](x,y) = (36,6)[/tex]

[tex]f(x) = \sqrt x[/tex] ----- the equation of the curve

Required

The slope of f(x)

The slope (m) is calculated using:

[tex]m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}[/tex]

[tex](x,y) = (36,6)[/tex] implies that:

[tex]a = 36; f(a) = 6[/tex]

So, we have:

[tex]m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}[/tex]

[tex]m = \lim_{h \to 0} \frac{f(36 + h) - 6}{h}[/tex]

If [tex]f(x) = \sqrt x[/tex]; then:

[tex]f(36 + h) = \sqrt{36 + h}[/tex]

So, we have:

[tex]m = \lim_{h \to 0} \frac{\sqrt{36 + h} - 6}{h}[/tex]

Multiply by: [tex]\sqrt{36 + h} + 6[/tex]

[tex]m = \lim_{h \to 0} \frac{(\sqrt{36 + h} - 6)(\sqrt{36 + h} + 6)}{h(\sqrt{36 + h} + 6)}[/tex]

Expand the numerator

[tex]m = \lim_{h \to 0} \frac{36 + h - 36}{h(\sqrt{36 + h} + 6)}[/tex]

Collect like terms

[tex]m = \lim_{h \to 0} \frac{36 - 36+ h }{h(\sqrt{36 + h} + 6)}[/tex]

[tex]m = \lim_{h \to 0} \frac{h }{h(\sqrt{36 + h} + 6)}[/tex]

Cancel out h

[tex]m = \lim_{h \to 0} \frac{1}{\sqrt{36 + h} + 6}[/tex]

[tex]h \to 0[/tex] implies that we substitute 0 for h;

So, we have:

[tex]m = \frac{1}{\sqrt{36 + 0} + 6}[/tex]

[tex]m = \frac{1}{\sqrt{36} + 6}[/tex]

[tex]m = \frac{1}{6 + 6}[/tex]

[tex]m = \frac{1}{12}[/tex]

Hence, the slope is 1/12

Answer: [tex]t\in [\dfrac{1}{4},2][/tex]

Step-by-step explanation:

Given

Inequality is [tex]4t^2\leq9t-2[/tex]

Taking variables one side

[tex]\Rightarrow 4t^2-9t+2\leq0\\\Rightarrow 4t^2-8t-t+2\leq0\\\Rightarrow 4t(t-2)-1(t-2)\leq0\\\Rightarrow (4t-1)(t-2)\leq0[/tex]

Using wavy curve method

[tex]t\in [\dfrac{1}{4},2][/tex]

Common difference: 6

First term: 7

Second term: 13

Third term: 19

Fourth term: 25

Fifth term: 31

I hope this is correct and helps!

Answer to the following question is as follows;

Number of term in AP (N) = 10

Step-by-step explanation:

Given:

Sum of arithmetic progression (Sn) = 340

First term of AP (a) = 7

Common difference of AP (d) = 6

Find;

Number of term in AP (N)

Computation:

Sn = [n/2][2a + (n-1)d]

340 = [n/2][2(7) + (n-1)6]

340 = [n/2][14 + 6n - 6]

680 = n[6n + 8]

6n² + 8n - 680

Using Quadratic Formula

n = 10

Number of term in AP (N) = 10

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step by step explanation:

[tex]\mathfrak{x}^{2}+{y}^{2}+16=0[/tex]

=[x2+16=0x26]

=[2x{y}^2{16}~0]

=[4×{y}^0{16}]

=[32x{y}^x]

Answer:

108 square metres

Step-by-step explanation:

A=√d square - l square

here

A = area

d= diagonal

l= length

Answer:

D

Step-by-step explanation:

There 2 ways to interpret this problem.

From the info given:

These two triangles are congruent by SSS and congruent triangles have congruent or equal side lengths so the answer have to be 1.

If the triangles are similar, the side lengths form a proportion of that

[tex] \frac{3}{3} = \frac{3}{3} [/tex]

So the ratio or scale factor is 1.

The scale factor in the figure is 1.

A scale factor in math is the ratio between corresponding measurements of an object and a representation of that object.

Given that two triangles, LMN and OPQ, we need to find the scale factor,

We can see triangles are congruent, and we know that

Two triangles are congruent, by the SSS congruence criterion, if they are similar and the scale factor happens to be 1,

Hence, the scale factor in the figure is 1.

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Answer:

f(6)=-42

f(0)=0

f(-1)=7

Step-by-step explanation:

since f(x) =-7x

it implies that f(6),f(0) and f(-1) are all equal to f(x)

which is equal to -7x .

so wherever you find x you out it in the

function

Answer:

(a): The conditional pmf of Y when X = 1

[tex]p_{Y|X}(0|1) = 0.2353[/tex]

[tex]p_{Y|X}(1|1) = 0.5882[/tex]

[tex]p_{Y|X}(2|1) = 0.1765[/tex]

(b): The conditional pmf of Y when X = 2

[tex]p_{Y|X}(0|2) = 0.0962[/tex]

[tex]p_{Y|X}(1|2) = 0.2692[/tex]

[tex]p_{Y|X}(2|2) = 0.6346[/tex]

(c): From (b) calculate P(Y<=1 | X =2)

[tex]P(Y\le1 | X =2) = 0.3654[/tex]

(d): The conditional pmf of X when Y = 2

[tex]p_{X|Y}(0|2) = 0.025[/tex]

[tex]p_{X|Y}(1|2) = 0.150[/tex]

[tex]p_{X|Y}(2|2) = 0.825[/tex]

Step-by-step explanation:

Given

The above table

Solving (a): The conditional pmf of Y when X = 1

This implies that we calculate

[tex]p_{Y|X}(0|1), p_{Y|X}(1|1), p_{Y|X}(2|1)[/tex]

So, we have:

[tex]p_{Y|X}(0|1) = \frac{p(y = 0\ n\ x = 1)}{p(x = 1)}[/tex]

Reading the data from the given table, the equation becomes

[tex]p_{Y|X}(0|1) = \frac{0.08}{0.08+0.20+0.06}[/tex]

[tex]p_{Y|X}(0|1) = \frac{0.08}{0.34}[/tex]

[tex]p_{Y|X}(0|1) = 0.2353[/tex]

Using the format of the above formula for the rest, we have:

[tex]p_{Y|X}(1|1) = \frac{0.20}{0.34}[/tex]

[tex]p_{Y|X}(1|1) = 0.5882[/tex]

[tex]p_{Y|X}(2|1) = \frac{0.06}{0.34}[/tex]

[tex]p_{Y|X}(2|1) = 0.1765[/tex]

Solving (b): The conditional pmf of Y when X = 2

This implies that we calculate

[tex]p_{Y|X}(0|2), p_{Y|X}(1|2), p_{Y|X}(2|2)[/tex]

So, we have:

[tex]p_{Y|X}(0|2) = \frac{p(y = 0\ n\ x = 2)}{p(x = 2)}[/tex]

Reading the data from the given table, the equation becomes

[tex]p_{Y|X}(0|2) = \frac{0.05}{0.05+0.14+0.33}[/tex]

[tex]p_{Y|X}(0|2) = \frac{0.05}{0.52}[/tex]

[tex]p_{Y|X}(0|2) = 0.0962[/tex]

Using the format of the above formula for the rest, we have:

[tex]p_{Y|X}(1|2) = \frac{0.14}{0.52}[/tex]

[tex]p_{Y|X}(1|2) = 0.2692[/tex]

[tex]p_{Y|X}(2|2) = \frac{0.33}{0.52}[/tex]

[tex]p_{Y|X}(2|2) = 0.6346[/tex]

Solving (c): From (b) calculate P(Y<=1 | X =2)

To do this, where Y = 0 or 1

So, we have:

[tex]P(Y\le1 | X =2) = P_{Y|X}(0|2) + P_{Y|X}(1|2)[/tex]

[tex]P(Y\le1 | X =2) = 0.0962 + 0.2692[/tex]

[tex]P(Y\le1 | X =2) = 0.3654[/tex]

Solving (d): The conditional pmf of X when Y = 2

This implies that we calculate

[tex]p_{X|Y}(0|2), p_{X|Y}(1|2), p_{X|Y}(2|2)[/tex]

So, we have:

[tex]p_{X|Y}(0|2) = \frac{p(x = 0\ n\ y = 2)}{p(y = 2)}[/tex]

Reading the data from the given table, the equation becomes

[tex]p_{X|Y}(0|2) = \frac{0.01}{0.01+0.06+0.33}[/tex]

[tex]p_{X|Y}(0|2) = \frac{0.01}{0.40}[/tex]

[tex]p_{X|Y}(0|2) = 0.025[/tex]

Using the format of the above formula for the rest, we have:

[tex]p_{X|Y}(1|2) = \frac{0.06}{0.40}[/tex]

[tex]p_{X|Y}(1|2) = 0.150[/tex]

[tex]p_{X|Y}(2|2) = \frac{0.33}{0.40}[/tex]

[tex]p_{X|Y}(2|2) = 0.825[/tex]

Step-by-step explanation:

here is the answer. Feel free to ask for more.

Hi!

√80 = √(16 • 5) = √(4² • 5) = 4√5

(1) Recall that

sin(x - y) = sin(x) cos(y) - cos(x) sin(y)

sin²(x) + cos²(x) = 1

Given that α lies in the third quadrant, and β lies in the fourth quadrant, we expect to have

• sin(α) < 0 and cos(α) < 0

• sin(β) < 0 and cos(β) > 0

Solve for cos(α) and sin(β) :

cos(α) = -√(1 - sin²(α)) = -3/5

sin(β) = -√(1 - cos²(β)) = -5/13

Then

sin(α - β) = sin(α) cos(β) - cos(α) sin(β) =  (-4/5) (12/13) - (-3/5) (-5/13)

==>   sin(α - β) = -63/65

(2) In the second identity listed above, multiplying through both sides by 1/cos²(x) gives another identity,

sin²(x)/cos²(x) + cos²(x)/cos²(x) = 1/cos²(x)

==>   tan²(x) + 1 = sec²(x)

Rewrite the equation as

3 sec²(x) tan(x) = 4 tan(x)

3 (tan²(x) + 1) tan(x) = 4 tan(x)

3 tan³(x) + 3 tan(x) = 4 tan(x)

3 tan³(x) - tan(x) = 0

tan(x) (3 tan²(x) - 1) = 0

Solve for x :

tan(x) = 0   or   3 tan²(x) - 1 = 0

tan(x) = 0   or   tan²(x) = 1/3

tan(x) = 0   or   tan(x) = ±√(1/3)

x = arctan(0) + nπ   or   x = arctan(1/√3) + nπ   or   x = arctan(-1/√3) + nπ

x = nπ   or   x = π/6 + nπ   or   x = -π/6 + nπ

where n is any integer. In the interval [0, 2π), we get the solutions

x = 0, π/6, 5π/6, π, 7π/6, 11π/6

(3) You only need to rewrite the first term:

[tex]\dfrac{\sin(x)}{1-\cos(x)} \times \dfrac{1+\cos(x)}{1+\cos(x)} = \dfrac{\sin(x)(1+\cos(x))}{1-\cos^2(x)} = \dfrac{\sin(x)(1+\cos(x)}{\sin^2(x)} = \dfrac{1+\cos(x)}{\sin(x)}[/tex]

Then

[tex]\dfrac{\sin(x)}{1-\cos(x)}+\dfrac{1-\cos(x)}{\sin(x)} = \dfrac{1+\cos(x)+1-\cos(x)}{\sin(x)}=\dfrac2{\sin(x)}[/tex]

Answer:

A and b Are in quadrant 2. F and D are in quadrant 1. F is in quadrant 3 and C is in quadrant 4

Step-by-step explanation:

each quadrant is in the boxes and the question is asking what is each  coordinate is in what quadrant