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pickupchik [31]
3 years ago
8

33. Given the proportion a/b=c/d solve for c​

Mathematics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

 \frac{ad}{b}

Step-by-step explanation:

Given proportion:

                      \frac{a}{b}   =  \frac{c}{d}

To solve this problem, make c the subject of the expression.

  Cross multiply;

                 ad  = cb

Now, divide both sides by b,

      \frac{ad}{b}   =  \frac{cb}{b}

          b  =  \frac{ad}{b}

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At her health club, Lauren uses a treadmill every 2 days and the weight machines every 8 days. She used a treadmill on March 2 a
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Answer:

Yes

Step-by-step explanation:

The Least common multiple (LCM) shows when the two over lap first.

3 0
3 years ago
Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)
DiKsa [7]

Answer:

C

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. y'=(3\sin x)'=3\cos x. Differentiate again to get y''=(3\cos x)'=-3\sin x, then y''+y=-3\sin x+3\sin x=0\neq 3\sin x so this choice of y doesn't solve the equation.

B. y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x and y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x, then y''+y=10\sin x+6\cos x\neq 3\sin x so y is not a solution

C. y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x hence y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x. Then y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x so y is a solution.

D.y'=-3\sin x and y''=-3\cos x, then y''+y=0 thus y isn't a solution

E. y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x hence y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x. Then y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x then y is not a solution.

3 0
3 years ago
Sara pays $2,120 in bills each month. Her monthly income is $2,960. What is Sara's disposable income?
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Saras disposable income is $840
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3 years ago
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A study conducted at a certain high school shows that 72% of its graduates enroll at a college. Find the probability that among
mixas84 [53]

Answer:

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

And we can use the probability mass function and we got:

P(X=0)=(4C0)(0.72)^0 (1-0.72)^{4-0}=0.00615

And replacing we got:

P(X \geq 1) = 1-0.00615 = 0.99385

Step-by-step explanation:

Let X the random variable of interest "number of graduates who enroll in college", on this case we now that:  

X \sim Binom(n=4, p=0.72)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

We want to find the following probability:

P(X \geq 1)

And we can use the complement rule and we got:

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

And we can use the probability mass function and we got:

P(X=0)=(4C0)(0.72)^0 (1-0.72)^{4-0}=0.00615

And replacing we got:

P(X \geq 1) = 1-0.00615 = 0.99385

5 0
4 years ago
please answer asap! A couple plans to have 5 children. The gender of each child is equally likely. Design a simulation involving
dimulka [17.4K]

Step-by-step explanation:

The gender of a child which is either a boy or a girl is determined by the XX-chromosomes, or XY-chromosomes.

Since the couple plan to have 5 children, the chance of a child being a boy is equal to the chance of it being a girl - the chances are 50/50.

What we do to achieve our aim is to run a simulation that would add an X or Y to an X for all 5 children.

Doing this 125 times, we obtain the number of trials we desire.

For each trial, we get for each child, C:

C1: X + (X or Y)

C2: X + (X or Y)

C3: X + (X or Y)

C4: X + (X or Y)

C5: X + (X or Y)

Since the chance of having an X is equal to the chance of having a Y, they equal probability, which is 0.5 for each.

6 0
3 years ago
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