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alekssr [168]
3 years ago
14

Brook paid $45 for a candle making course. She spent $106 for wax and supplies , but later returned one $8 candle mold. Write an

d find out the value of an integer expression to show the change in the amount of money she has . If brook sells her candles for $9 each how many will she have to sell before she makes a profit
Mathematics
1 answer:
postnew [5]3 years ago
7 0
She spent $143 after returns and to make a profit at $9 dollars per candle, she would need to sell 16 Candles. molly ringwald tho.
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Find the LCM of 6 and 8 and list the multiples <br>​
Hitman42 [59]

Answer:

24

Step-by-step explanation:

6x4

8x3

8 0
3 years ago
You want to test the effect of light on the distribution of Artemia in your 35cm long testing chamber. If you measure the light
Dvinal [7]

Answer:

64.3 lumens

Step-by-step explanation:

Data provided in the question:

Length of the testing chamber = 35 cm

Light at one end = 10 lumens

Light at other end = 200 lumens

now,

since the variation is linear,

therefore, the gradient of the variation

= \frac{200-10}{35}

= 5.43

also,

the darkest end is the end with 10 lumens

therefore,

The lumens at 10 cm from the darkest end

= 10 + 5.43(10)

= 10 + 54.3

= 64.3 lumens

6 0
3 years ago
What does point A represent in this box plot?
Zigmanuir [339]
Point A means the smallest value (and = 4)

answer
<span>C. the smallest value</span>
3 0
3 years ago
Read 2 more answers
Graph the line y=−3x+b if it is known that the graph goes through point:
Gemiola [76]

Answer:

А(0/6;-2) и B(-5.6;17)

Step-by-step explanation:

(-2;4) поставь координаты место (x;y)

Например: 4=-3*(-2)+b, найдите b, а затем нарисуйте график.

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8 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
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