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ale4655 [162]
3 years ago
15

Someone please help me

Mathematics
1 answer:
raketka [301]3 years ago
6 0
The answer is $303.

The debt coming from the fee is stable, it is $15.

There are 9 tickets, which costs $32 per ticket, the debt from them is:
9 \times 32 = 288
Add them together:
288 + 15 = 303
You might be interested in
Find solution to initial value problem: y'' +4y' +7y = 0; y(0) = 1, y'(0) = -2
rusak2 [61]
Try this option, modify design according to local requirements.

4 0
3 years ago
A metal hollow bar whose cross section and dimension are shown below weighs 8x10^3 kg/m^3 and measure 2m in length ..determine t
devlian [24]
1) We calculate the volume of a metal bar (without the hole).

volume=area of hexagon x length
area of hexagon=(3√3 Side²)/2=(3√3(60 cm)²) / 2=9353.07 cm²
9353.07 cm²=9353.07 cm²(1 m² / 10000 cm²)=0.935 m²

Volume=(0.935 m²)(2 m)=1.871 m³

2) we calculate the volume of the parallelepiped

Volume of a parallelepiped= area of the section  x length
area of the section=side²=(40 cm)²=1600 cm²
1600 cm²=(1600 cm²)(1 m² / 10000 cm²=0.16 m²
Volume of a parallelepiped=(0.16 m²)(2 m)=0.32 m³

3) we calculate the volume of a metal hollow bar:
volume of a metal hollow bar=volume of a metal bar -  volume of a parallelepiped

Volume of a metal hollow bar=1.871 m³ - 0.32 m³=1.551 m³

4) we calculate the mass of the metal bar

density=mass/ volume  ⇒ mass=density *volume

Data:
density=8.10³ kg/m³
volume=1.551 m³

mass=(8x10³ Kg/m³ )12. * (1.551 m³)=12.408x10³ Kg

answer: The mas of the metal bar is 12.408x10³ kg  or   12408 kg  


4 0
3 years ago
A slitter assembly contains 48 blades. Five blades are selected at random and evaluated each day of sharpness. If any dull blade
Alex73 [517]

Answer:

Part a

The probability that assembly is replaced the first day is 0.7069.

Part b

The probability that assembly is replaced no replaced until the third day of evaluation is 0.0607.

Part c

The probability that the assembly is not replaced until the third day of evaluation is 0.2811.

Step-by-step explanation:

Hypergeometric Distribution: A random variable x that represents number of success of the n trails without replacement and M represents number of success of the N trails without replacement is termed as the hypergeometric distribution. Moreover, it consists of fixed number of trails and also the two possible outcomes for each trail.

It occurs when there is finite population and samples are taken without replacement.

The probability distribution of the hyper geometric is,

P(x,N,n,M)=\frac{(\limits^M_x)(\imits^{N-M}_{n-x})}{(\limits^N_n)}

Here x is the success in the sample of n trails, N represents the total population, n represents the random sample from the total population and M represents the success in the population.

Probability that at least one of the trail is succeed is,

P(x\geq1)=1-P(x

(a)

Compute the probability that the assembly is replaced the first day.

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly N = 48.

Number of blades selected at random from the assembly  n= 5

Number of blades in an assembly dull is M  = 10.

The probability mass function is,

P(X=x)=\frac{[\limits^M_x][\limits^{N-M}_{n-x}]}{[\limits^N_n]};x=0,1,2,...,n\\\\=\frac{[\limits^{10}_x][\limits^{48-10}_{5-x}]}{[\limits^{48}_5]}

The probability that assembly is replaced the first day means the probability that at least one blade is dull is,

P(x\geq 1)=1- P(x

(b)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly  N = 48

Number of blades selected at random from the assembly  N = 5

Number of blades in an assembly dull is  M = 10

From the information,

The probability that assembly is replaced (P)  is 0.7069.

The probability that assembly is not replaced is (Q)  is,

q=1-p\\= 1-0.7069= 0.2931

The geometric probability mass function is,

P(X = x)= q^{x-1} p; x =1,2,....=(0.2931)^{x-1}(0.7069)

The probability that assembly is replaced no replaced until the third day of evaluation is,

P(X = 3)=(0.2931)^{3-1}(0.7069)\\=(0.2931)^2(0.7069)= 0.0607

(c)

From the given information,

Let x be number of blades dull in the assembly are replaced.

Total number of blades in the assembly   N = 48

Number of blades selected at random from the assembly  n = 5

Suppose that on the first day of the evaluation two of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 2.

P(x=0)=\frac{(\limits^2_0)(\limits^{48-2}_{5-0})}{\limits^{48}_5}\\\\=\frac{(\limits^{46}_5)}{(\limits^{48}_5)}\\\\= 0.8005

Suppose that on the second day of the evaluation six of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M  = 6.

P(x=0)=\frac{(\limits^6_0)(\limits^{48-6}_{5-0})}{(\limits^{48}_5)}\\\\=\frac{(\limits^{42}_5}{(\limits^{48}_5)}\\\\= 0.4968

Suppose that on the third day of the evaluation ten of the blades are dull then the probability that the assembly is not replaced is,

Here, number of blades in an assembly dull is M

= 10.

P(x\geq 1)=1- P(x

 

The probability that the assembly is not replaced until the third day of evaluation is,

P(The assembly is not replaced until the third day)=P(The assembly is not replaced first day) x P(The assembly is not replaced second day) x P(The assembly is replaced third day)

=(0.8005)(0.4968)(0.7069)= 0.2811

5 0
3 years ago
BE QUICK! WILL GIVE BRAINLIEST TO CORRECT ANSWER
storchak [24]

Answer:

$7.5

Step-by-step explanation:

3/.4

6 0
3 years ago
Draw a right triangle with side lengths of 3, 4, and 5 units. <br> and answer the problem please :')
11111nata11111 [884]

Answer:

Image below

Step-by-step explanation:

<em>Given: Side lengths of a right triangle 3,4 and 5 units. </em>

<em> To draw: A right triangle with the given side length. </em>

<em> Solution: </em>

<em> We know, in a right angle triangle hypotenuse is the longest side and satisfying Pythagoras theorem. </em>

<em> From the given side length, </em>

<em> Hypotenuse = 5 unit </em>

<em> We can take any of the base and perpendicular. </em>

<em> Let, Base = 3 unit </em>

<em> Perpendicular = 4 unit. </em>

<em> It a right-angle triangle with a hypotenuse 5 unit. </em>

<em> Now we draw a right angle triangle taking in the first 3 base and 4 perpendicular and second 3 perpendicular and 4 bases.</em>

8 0
2 years ago
Read 2 more answers
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