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3 years ago

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1. 15% of the toddlers in a preschool class drink water with their lunch. How many toddlers are in the class if 3 drink water wi

th their lunch?
(a) Write a percent equation for the situation.
(b) Solve the problem. Show your work.
plz help tysm <3

1 answer:

USPshnik [31]3 years ago

8 0

Answer:

20

Step-by-step explanation:

Let t represent the number of toddlers in the class. Then 15% of t = 3.

In other terms, 0.15t = 3, and t = 3/0.15 = 20.

There are 20 toddlers in the class.

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A taxi driver had 27 fares to and from the airport last Monday. The price for a ride to the airport is $14, and the price for a

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Answer:

(15, 12)

Step-by-step explanation:

Let's generate two systems of equations that fit this scenario.

Number of trips to the airport = x

Number of trips from the airport = y

Total number of trips to and from the airport = 27

Thus:

$x + y = 27$ => equation 1.

Total price for trips to the Airport = 14*x = 14x

Total price of trips from the airport = 7*y = 7y

Total collected for the day = $294

Thus:

$14x + 7y = 294$ => equation 2.

Multiply equation 1 by 7, and multiply equation 2 by 1 to make both equations equivalent.

7 × $x + y = 27$

1 × $14x + 7y = 294$

Thus:

$7x + 7y = 189$ => equation 3

$14x + 7y = 294$ => equation 4

Subtract equation 4 from equation 3

-7x = -105

Divide both sides by -7

x = 15

Substitute x = 15 in equation 1

$x + y = 27$

$15 + y = 27$

Subtract both sides by 15

$y = 27 - 15$

$y = 12$

The ordered pair would be (15, 12)

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2 years ago

Use the number line to find the equivalent fraction

kifflom [539]

I think that the Equivalent fraction is 6/8

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3 years ago

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A slide 15 feet long makes an angle of 330 with its vertical ladder. To the

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Answer:

13ft

Step-by-step explanation:

Kindly find attached a rough draft of the situation.

Step one:

Given data

The length of the slide represents the

Hypotenuse of the situation on the rough sketch

Angle =33°

Required

The height of the ladder which is the adjacent of the rough sketch represented by x

Step two:

Applying SOH CAH TOA

Cos θ= adj/hyp

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Cross multiplying

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2 years ago

How to write the equation in standard form

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Answer:

Step-by-step explayesnation:

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3 years ago

Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p

Ymorist [56]

Answer:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

$A\sim Uniform(1 ,7)$

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

$B\sim Exp(\lambda=\frac{1}{30 min})$

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

$E(A +B)$

Since the two random variables are assumed independent, then we have this

$E(A+B) = E(A)+E(B)$

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by $E(X) =\frac{a+b}{2}$ where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

$E(A)=\frac{1+7}{2}=4 hours$

The expected value for the exponential distirbution is given by :

$E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx$

If we use the substitution $y=\lambda x$ we have this:

$E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}$

Where X represent the random variable and $\lambda$ the parameter. If we apply this formula to our case we got:

$E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min$

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

$E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours$

And in order to find the variance for the random variable A+B we can find the individual variances:

$Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2$

$Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2$

We have the following property:

$Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)$

Since we have independnet variable the Cov(A,B)=0, so then:

$Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2$

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3 years ago