Mike forgot the last three digits of his friend's telephone number. He only remembers that the digits are not the same. What is the probability he dials the correct telephone number on the first try

Answer:

86 boxes

Step-by-step explanation:

trust me its right

Answer:

angle M = 60

angle Q = 70

Step-by-step explanation:

M 180/3 = 60

Q 180-40 = 140/2 = 70

Answer:

Period: [tex]\pi[/tex]

Phase shift: n

Step-by-step explanation:

Tangent function:

Has the following format:

[tex]f(x) = \tan{ax - n}[/tex]

In which the period is [tex]\frac{\pi}{x}[/tex] and the phase shift is n.

In this question:

[tex]f(x) = -4\tan{(x-n)}[/tex]

[tex]a = 1[/tex], and thus, the period is [tex]\pi[/tex], with a phase shift of n.

Answer:

i believe the pre-tax subtotal would be 1890.69

Step-by-step explanation:

the 2,033 represents 100%. to remove that 7% you would do

.93 • 2,033 which gives you 1890.69

Answer:

0.2x+5>2

Step-by-step explanation:

0.2 is the same as 2/10;

(2/10)x>2-5

(2/10)x>-3

2x>-30

X>-15( since -15 is lesser than -14,-13,-12 and so on. the sign should be >

ಠ_ಠ (눈‸눈) (⌐■-■)

(ب_ب) ¯\_ಠ_ಠ_/¯

Answer:

20$ i think

Step-by-step explanation:

A fraction is a way to describe a part of a whole. If a car is parked in the garage for 5 hours and 30 minutes the charge will be $10.

A fraction is a way to describe a part of a whole. such as the fraction ¼ can be described as 0.25.

Given that for the first hour that a car is parked in the garage, there is no charge. After the first hour, for the next two hours that a car is parked in the garage, there is a $5 charge. After the third hour, the garage charges $2 for each additional hour that the car is parked in the garage.

1.) Since the car is parked for 30 minutes only.

Charge = $0

Hence, No charge will be charged for 30 minutes.

2.) If a car is parked in the garage for 2 hours and 30 minutes.

Charge for 1 hour = $0

Charge for the next 1 hour and 30 minutes = $5

Hence, the charge will be $5.

3.)  If a car is parked in the garage for 5 hours.

Charge for 1 hour = $0

Charge for the next 2 hour = $5

Charge for the next 2 hour = 2×$2 = $4

Charge = $5 + $4 = $9

Hence, the charge will be $9.

4.) If a car is parked in the garage for 5 hours and 30 minutes.

Charge for 1 hour = $0

Charge for the next 2 hour = $5

Charge for the next 2 hour 30 minutes = 2.5 ×$2 = $5

Charge = $5 + $5 = $10

Hence, the charge will be $10.

Learn more about Fraction:

https://brainly.com/question/1301963

#SPJ2

9514 1404 393

Answer:

  A. perhaps 0 to 7 days

  B. the radius in mm at the start of the study

  C. 0.114 mm/day

Step-by-step explanation:

A. It is always problematic to determine the reasonable domain for an exponential growth function. You can always limit the domain to the length of time for which the function is said to model the growth. It is difficult to say how much beyond that time period the exponential growth can be extrapolated, as most systems run into limits to growth.

Here, the base of the exponential term is 1.01 = 1 + 1%. This tells you the growth rate is 1% per day. The study concludes when the radius was 11.79/11.00 = 1.071818... ≈ 1 + 7.2% times the original size. That is, the study lasted approximately 7 days.

The question in part C has you look at the size on day 7, which is apparently the last day of the study. It is not clear that the model is at all useful beyond the end of the period it is intended to describe.

A reasonable domain for the growth function is 0 to 7 days.

__

B. The function gives the radius of the algae after d days. When d=0 (the y-intercept), the function gives the radius of the algae after 0 days. That is the meaning of the y-intercept is the initial radius of the algae (at the beginning of the study).

The y-intercept is always the "initial value", the value when the independent variable is zero. You have to look at the function definition to see what it is the initial value of.

__

C. The average rate of change is the difference in function values divided by the difference in time between them. Here, the attached table tells us that is ...

  (f(7) -f(2))/(7 -2) = (11.793 -11.221)/5 = 0.572/5 = 0.1144 . . . mm/day

The units of the "average rate of change" are the units of the slope of the curve on the graph: the y-axis units divided by the x-axis units. Here, that is mm/day.

Answer:

(A) [tex]5\frac{1}{4}*4\frac{1}{5}[/tex]

Step-by-step explanation:

The area of a parallelogram is the same as the area of a rectangle which is A=bh where b is the base and h is the height. Therefore, Erica can use the expression [tex]5\frac{1}{4}*4\frac{1}{5}[/tex] to find the area of the parallelogram.

Answer:

To be able to approximate the sampling distribution with a normal model, it is needed that [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], and both conditions are satisfied in this problem.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they will make payments on time, or they won't. The probability of a person making the payment on time is independent of any other person, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

Probability of exactly x successes on n repeated trials, with p probability.

The sampling distribution can be approximated to a normal model if:

[tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex]

Based on past experience, a bank believes that 8.9 % of the people who receive loans will not make payments on time.

This means that [tex]p = 0.089[/tex]

The bank has recently approved 220 loans.

This means that [tex]n = 220[/tex]

What must be true to be able to approximate the sampling distribution with a normal model?

[tex]np = 220*0.089 = 19.58 \geq 10[/tex]

[tex]n(1-p) = 220*0.911 = 200.42 \geq 10[/tex]

To be able to approximate the sampling distribution with a normal model, it is needed that [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex], and both conditions are satisfied in this problem.

Answer:

18.0167≤x≤21.9833

Step-by-step explanation:

Given the following

sample size n = 10

standard deviation s = 3.2

Sample mean = 20

z-score at 95% = 1.96

Confidence Interval = x ± z×s/√n

Confidence Interval = 20 ± 1.96×3.2/√10

Confidence Interval = 20 ± (1.96×3.2/3.16)

Confidence Interval = 20 ± (1.96×1.0119)

Confidence Interval = 20 ±  1.9833

CI = {20-1.9833, 20+1.9833}

CI = {18.0167, 21.9833}

Hence the required confidence interval is 18.0167≤x≤21.9833

Answer and Step-by-step explanation:

This is for your other question in case you don't see it.

1. 2 pairs (aka 4 sides) of Opposite Sides are equal

2. AB and DC are parallel, and AD and BC are parallel  

3. Angles BDC and ACD are equal, angles DAC and DBC are equal, the angles ADB and BCD are equal, angles CAB and DBA are equal

4. 4 right angles

5. AB and DC are equal, and AD and BC are equal

6. 4 Triangles

7. False

8. Diagonals of a rectangle.

#teamtrees #PAW (Plant And Water)

Answer:

the answer is ( -8,-8 )

Step-by-step explanation:

the first point is in -8 and the second point is also in -8 , therefore the answer is ( -8,-8 )

Answer:

81=1/2h×9,

81=1/18h

1458h=1

h=1/1458

Answer:

h=1/ 1458

hope it is helpful to you

Answer:

Equation: 4x + 30 + 2x + 6 = 180

Answer: x = 24

Step-by-step explanation:

The sum of the measures of supplementary angles is 180 deg.

Equation:

4x + 30 + 2x + 6 = 180

Solution:

4x + 30 + 2x + 6 = 180

Add like terms on the left side.

6x + 36 = 180

Subtract 36 from both sides.

6x = 144

x = 24

Answer:

X=24

Step-by-step explanation:

Supplementary angles = 180°

4x+30+2x+6=180

Combine like terms> 4x+2x=6x

Add: 30+6=36

6x+36=180.

Subtract 36 on both sides. > 36-36=0. 180-36=144.

Drop what you have left> 6x   144

Divide by 6. > 6/6= 1. 144/6=24.

X=24

Answer:

1.

[tex]midpoint = ( \frac{ - 4 + 6}{2} , \: \frac{ - 12 + 8}{2} ) \\ = (1, \: - 2)[/tex]

2.

[tex]distance = \sqrt{ {( - 4 - 6)}^{2} + {( - 12 - 8)}^{2} } \\ = \sqrt{500} \\ = 22.4 \: units[/tex]

Answer:

(0.5X) - 3 = Distance from the finish line

Step-by-step explanation:

Given that a student is running a 3-kilometer race, and runs 1 kilometer every 2 minutes, to determine the function that describes the distance from the finish line after X minutes, the following calculation must be performed:

1 = kilometers for every 2 minutes

X = every number of minutes

1/2 = 0.5 = kilometers per minute

(0.5X) - 3 = Distance from the finish line

Thus, if the student runs for 4 minutes, the equation would operate as follows:

0.5 x 4 - 3 = X

2 - 3 = X

-1 = X

Answer:

The maximum volume of the box is:

[tex]V =\frac{5}{3}\sqrt{\frac{5}{3}}[/tex]

Step-by-step explanation:

Given

[tex]Surface\ Area = 10m^2[/tex]

Required

The maximum volume of the box

Let

[tex]a \to base\ dimension[/tex]

[tex]b \to height[/tex]

The surface area of the box is:

[tex]Surface\ Area = 2(a*a + a*b + a*b)[/tex]

[tex]Surface\ Area = 2(a^2 + ab + ab)[/tex]

[tex]Surface\ Area = 2(a^2 + 2ab)[/tex]

So, we have:

[tex]2(a^2 + 2ab) = 10[/tex]

Divide both sides by 2

[tex]a^2 + 2ab = 5[/tex]

Make b the subject

[tex]2ab = 5 -a^2[/tex]

[tex]b = \frac{5 -a^2}{2a}[/tex]

The volume of the box is:

[tex]V = a*a*b[/tex]

[tex]V = a^2b[/tex]

Substitute: [tex]b = \frac{5 -a^2}{2a}[/tex]

[tex]V = a^2*\frac{5 - a^2}{2a}[/tex]

[tex]V = a*\frac{5 - a^2}{2}[/tex]

[tex]V = \frac{5a - a^3}{2}[/tex]

Spit

[tex]V = \frac{5a}{2} - \frac{a^3}{2}[/tex]

Differentiate V with respect to a

[tex]V' = \frac{5}{2} -3 * \frac{a^2}{2}[/tex]

[tex]V' = \frac{5}{2} -\frac{3a^2}{2}[/tex]

Set [tex]V' =0[/tex] to calculate a

[tex]0 = \frac{5}{2} -\frac{3a^2}{2}[/tex]

Collect like terms

[tex]\frac{3a^2}{2} = \frac{5}{2}[/tex]

Multiply both sides by 2

[tex]3a^2= 5[/tex]

Solve for a

[tex]a^2= \frac{5}{3}[/tex]

[tex]a= \sqrt{\frac{5}{3}}[/tex]

Recall that:

[tex]b = \frac{5 -a^2}{2a}[/tex]

[tex]b = \frac{5 -(\sqrt{\frac{5}{3}})^2}{2*\sqrt{\frac{5}{3}}}[/tex]

[tex]b = \frac{5 -\frac{5}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]

[tex]b = \frac{\frac{15 - 5}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]

[tex]b = \frac{\frac{10}{3}}{2*\sqrt{\frac{5}{3}}}[/tex]

[tex]b = \frac{\frac{5}{3}}{\sqrt{\frac{5}{3}}}[/tex]

Apply law of indices

[tex]b = (\frac{5}{3})^{1 - \frac{1}{2}}[/tex]

[tex]b = (\frac{5}{3})^{\frac{1}{2}}[/tex]

[tex]b = \sqrt{\frac{5}{3}}[/tex]

So:

[tex]V = a^2b[/tex]

[tex]V =\sqrt{(\frac{5}{3})^2} * \sqrt{\frac{5}{3}}[/tex]

[tex]V =\frac{5}{3} * \sqrt{\frac{5}{3}}[/tex]

[tex]V =\frac{5}{3}\sqrt{\frac{5}{3}}[/tex]

The maximum volume of the box which has a 10 m² surface area is given below.

[tex]\rm V_{max} = \dfrac{5}{3} *\sqrt{\dfrac{5}{2}}[/tex]

The rate of change of a function with respect to the variable is called differentiation. It can be increasing or decreasing.

We want to construct a box with a square base and we currently only have 10 m² of material to use in the construction of the box.

The surface area = 10 m²

Let a be the base length and b be the height of the box.

Surface area = 2(a² + 2ab)

  2(a² + 2ab) = 10

      a² + 2ab = 5

Then the value of b will be

[tex]\rm b = \dfrac{5-a^2}{2a}[/tex]

The volume of the box is given as

V = a²b

Then we have

[tex]\rm V = \dfrac{5-a^2 }{2a}* a^2\\\\V = \dfrac{5a - a^3}{2}\\\\V = \dfrac{5a}{2} - \dfrac{a^3}{2}[/tex]

Differentiate the equation with respect to a, and put it equal to zero for the volume to be maximum.

[tex]\begin{aligned} \dfrac{dV}{da} &= \dfrac{d}{da} ( \dfrac{5a}{2} - \dfrac{a^3}{2} ) \\\\\dfrac{dV}{da} &= 0 \\\\\dfrac{5}{2} - \dfrac{3a^2 }{2} &= 0\\\\a &= \sqrt{\dfrac{5}{2}} \end{aligned}[/tex]

Then the value of b will be

[tex]b = \dfrac{5-\sqrt{\dfrac{5}{2}} }{2*\sqrt{\dfrac{5}{2}} }\\\\\\b = \sqrt{\dfrac{5}{2}}[/tex]

Then the volume will be

[tex]\rm V = (\sqrt{\dfrac{5}{2}} )^2*\sqrt{\dfrac{5}{2}} \\\\V = \dfrac{5}{3} *\sqrt{\dfrac{5}{2}}[/tex]

More about the differentiation link is given below.

https://brainly.com/question/24062595

Answer:

There are 46 rows with 63 seats in each row

Step-by-step explanation:

I started looking for a whole number dividing seats and rows to make up the two pieces we need to multiply. I started backward from 70 (lucky guess) and then worked my way down to 63 and 46.

Now I was also looking for something more elegant in an algabraic formula and I stated with x being the number of rows and the seats being (x=17)

so I started with X(X+17)=2898 but that fif sot pan out other than to take me to x squared +17 = 2898 - subtract 17 from each side xsquared equals 2916

square root of 2916 is 54 which started my searching for a random number.

I got lucky

Answer:

Step-by-step explanation:

y=[tex]\frac{3}{2} x\\[/tex]-3

Answer:

630 feets

Step-by-step explanation:

From the triangle attached :

Using trigonometry, the height h, which is the height of climber to your eye level :

Tan θ = opposite / Adjacent

Tan 32 = h / 1000

h = tan 32 * 1000

h = 0.6248693 * 1000

h = 624.86935

Height from the ground :

624.86935 + 5.5

= 630.369 feets

= 630 feet

Answer:

[tex]B' = (-3,-3)[/tex]

[tex]C' = (-4.5,-7.5)[/tex]

Step-by-step explanation:

Given

[tex]k = 1.5[/tex]

[tex]B = (-2,2)[/tex]

[tex]C =(-3,-5)[/tex]

Required

B' and C'

This is calculated as;

[tex]B' = k * B[/tex]

[tex]B' = 1.5 * (-2,-2)[/tex]

[tex]B' = (-3,-3)[/tex]

and

[tex]C' =k * C[/tex]

[tex]C' = 1.5 * (-3,-5)[/tex]

[tex]C' = (-4.5,-7.5)[/tex]

243 = the pages read on Saturday

243 - 53 = 190 = the pages read on Sunday

190 + 243 =  433 = total pages read

Answer = 433

Answer:

3.75

Step-by-step explanation:

DF = 6/24 × 15 = 3.75

________________

Answer:

The length of the three sides [tex]5, \sqrt{58} , \sqrt{65}[/tex]

The triangle is not a right triangle

Step-by-step explanation:

A = (3, 2) , B = ( 6, 9) , C = (10, 6)

Find the lengths using distance formula.

[tex]distance = \sqrt{(x_2 -x_1)^2 + (y_2 - y_1)^2}[/tex]

[tex]AB = \sqrt{(6-3)^2 + (9-2)^2} = \sqrt{9 + 49 } = \sqrt{58}[/tex]

[tex]BC = \sqrt{(10-6)^2 + (6-9)^2} = \sqrt{16 + 9} = \sqrt{25} = 5[/tex]

[tex]AC = \sqrt{(3-10)^2+(2-6)^2} = \sqrt{49 + 16} = \sqrt{65}[/tex]

Using Pythagoras theorem :

[tex](Longer \ side)^2 = sum \ of \ square \ of \ two \ other \ sides[/tex]

Longest side is AC . So we will check if it satisfies Pythagoras theorem :

[tex]\sqrt{65} = \sqrt{58} + 5^2\\65 = 58 + 25\\[/tex]

65 ≠ 58 + 25

So the sides does not satisfy Pythagoras theorem. Hence the triangle is not a right triangle.

No they aren't

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