how many square tiles with sides 1 ft long on each edge would be needed to the floor of an 8-by-13ft room

Isabella has twice as much money invested at 4% as she has invested at 3%, and she has $500 more invested at 5% than she has at

adell [148]

Invested at 3% = x
invested at 4% = 2x
invested at 5% = x+500

0.03x + 0.04(2x) + 0.05(x+500) = 2025
0.03x + 0.08x +0.05x +25 = 2025
0.16x = 2000
x= 12,500
4% = 2x = 12,500 * 2 = $25,000

3 0

3 years ago

what is the correct upper and lower control limits for 8.42 pounds and the accectable range for plus or minus .65 pounds

sweet [91]

Answer:

9.07 and 7.77 respectively.

Step-by-step explanation:

To determine the upper control limit, add 0.65 lb to 8.42 lb, obtaining 9.07 lb.

To determine the lower one, subtract 0.65 lb from 8.42 lb, obtaining 7.77 lb.

7 0

4 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

4 years ago

Plz plz plz help help❤️ With atleast some

RSB [31]

Answer:

the tables have a pattern of consistancy

Step-by-step explanation:

7 0

4 years ago

Math question down below

Kipish [7]

Answer: 6928

----------------------------------------------------------

Explanation:

We have two areas we need to find: The area of the trapezoid and the area of the rectangle. Let's call these areas A1 and A2.

Area of Trapezoid = (height)*(base1+base2)/2
A1 = h*(b1+b2)/2
A1 = 80*(150+100)/2
A1 = 80*250/2
A1 = 20000
A1 = 10000

Area of Rectangle = (length)*(width)
A2 = L*W
A2 = 48*64
A2 = 3072

Subtract the two areas (A1-A2) to get the difference D
D = A1 - A2
D = 10000 - 3072
D = 6928

This difference D is exactly equal to the shaded area.

3 0

3 years ago