THIS PROBLEM IS WORTH 40 POINTS

Solve for x and verify:3(x+3)-2(x-1)=5(x-5)

Gekata [30.6K]

Answer:

x=9

Step-by-step explanation:

Distribute:

3x+9-2x+2=5x-25

Combine like terms:

x+11=5x-25

Solve:

x-x+11=5x-x-25

11=4x-25

11+25=4x-25+25

36=4x

Divide by 9:

x=9.

Hope this helps :D

7 0

3 years ago

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Last year, Jason's average in math was a 72. This year, his average is a 90. What is the percent increase in Jason's grade?

slamgirl [31]

Answer:

20% increase

Step-by-step explanation:

All you have to do is divide 72 by 90n and it will give you .8, move the decimal two places to the right to give you 80 and subtract 80 from 100 to get 20

7 0

2 years ago

4Brainliest

Maurinko [17]

The correct answer is:

$5.75

2.3% of 250 is 5.75.

5 0

3 years ago

Read 2 more answers

Water pressure increases 0.44 pounds per square inch (0.44 psi) with each increase of one foot in depth below sea level. Identif

prohojiy [21]

Answer:

Depth below sea level is the independent quantity,

Water pressure is the dependent quantity

Step-by-step explanation:

An independent quantity is a variable that can be changed in an experiment. While, dependent quantity results from the independent quantity or we can say, that depends upon the independent quantity.

Here,

The water pressure increases 0.44 pounds per square inch (0.44 psi) with each increase of one foot in depth below sea level,

So, for measuring the water pressure we took depth below sea level as a variable,

⇒ Depth below sea level is the independent quantity,

While, with increasing depth by 1 foot the pressure is also increase by 0.44 pounds per square inches ⇒ pressure depends upon the depth

⇒ Water pressure is the dependent quantity.

3 0

3 years ago

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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