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3 years ago

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Loren drove 200 miles at a certain rate, and his wife, Lois, drove 100 miles at a rate 10 mph slower. If Loren had driven for th

e entire trip, they would have arrived 30 minutes sooner. What was Loren's rate? 45 mph 50 mph 55 mph

2 answers:

Marta_Voda [28]3 years ago

6 0

See the attached picture

Marysya12 [62]3 years ago

6 0

Answer:

50 mph

Step-by-step explanation:

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How do you multiply mixed numbers with fractions

Vaselesa [24]

Answer:

Step-by-step explanation:

You first turn the mixed number into an improper fraction by mutliplying the denominator by the whole number then adding the numerator to the product. Make sure to keep the denominator the same and just put the sum as the numerator. Then, you multiply the fractions and simplify your answer! Hope this helped!

4 0

3 years ago

Read 2 more answers

The product of a binomial and a trinomial is x3+3x2-x+2x2+6x-2 Which expression is equivalent to this

solong [7]

The answer is the first one.
Explanation:
X^3 stays the same because there are no other cubed numbers in the problem
Next you combine the x^2s
The x^2s are +3x^2 and +2x^2
Since they are both positive, you add them: 3x^2 + 2x^2 = 5x^2
Next you do the x values
-x and +6x, also known as 6x - x = 5x
Lastly, you just add in the -2 and get:
X^3 + 5x^2 + 5x - 2

8 0

2 years ago

Carrie has 40 more nickels than Joan has dimes. They both have the same amount of money. How many coins does each girl have? Wha

LuckyWell [14K]

Answer:

Carrie and Joan have 80 nickels and 40 dimes, respectively, and each one has an amount of money of 4 US dollars.

Step-by-step explanation:

Let be $x$ and $y$ the amounts of nickels and dimes that Carrie and Joan have, as $c$ the total amount of money that both Carrie and Joan have. A nickel is a five cent coin and a dime is a ten cent coin.

The following equations are constructed after a careful reading on statement:

Carrie's amount of money:

$0.05\cdot x = c$ (Eq. 1)

Joan's amount of money:

$0.10\cdot y = c$ (Eq. 2)

Relation between amounts of coins:

$x = y+40$ (Eq. 3)

First we eliminate $c$ by equalizing (Eq. 1) and (Eq. 2):

$0.05\cdot x = 0.10\cdot y$

Then, we reduce the resulting formula by (Eq. 3):

$0.05\cdot (y+40) = 0.10\cdot y$

$y+40 = 2\cdot y$

$y = 40$

And rest of variable are now determined:

$x = 40+40$

$x = 80$

$c = 0.05\cdot (80)$

$c = 4$

Carrie and Joan have 80 nickels and 40 dimes, respectively, and each one has an amount of money of 4 US dollars.

5 0

3 years ago

A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t

julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\lambda$ is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with $\mu = \lambda, \sigma = \sqrt{\lambda}$ , if $\lambda>10$ .

Poisson variable with the mean 3

This means that $\lambda= 3$ .

(a) At least 3 in a week.

This is $P(X \geq 3)$ . So

$P(X \geq 3) = 1 - P(X < 3)$

In which:

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

Then

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768$

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

In which

$P(X = x) = \frac{e^{-\lambda}*\lambda^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-3}*3^{0}}{(0)!} = 0.0498$

$P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494$

$P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240$

$P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240$

$P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680$

$P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008$

$P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504$

$P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216$

Then

$P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988$

0.988 = 98.8% probability that the shop sells at most 7 in a week.

(c) More than 20 in a month (4 weeks).

4 weeks, so:

$\mu = \lambda = 4(3) = 12$

$\sigma = \sqrt{\lambda} = \sqrt{12}$

The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{20 - 12}{\sqrt{12}}$

$Z = 2.31$

$Z = 2.31$ has a p-value of 0.9896.

1 - 0.9896 = 0.0104

0.0104 = 1.04% probability that the shop sells more than 20 in a month.

5 0

2 years ago

A rectangle has a length 10 more than its width. If the width is increased by 8 and the length by 4, the resulting rectangle has

labwork [276]

The equation to model the above scenario is $x^{2}$ +22x - 23 = 0

The perimeter of the expanded rectangle is 48 units

<h3>What is a rectangle?</h3>

A rectangle is a quadrilateral with its 4 angles 90°

Analysis:

First rectangle:

length = 10 + x

width = x

Second rectangle:

length = x + 14

width = x + 8

Area of expanded rectangle = 135 square unit

(x+8)(x+14) = 135

$x^{2}$ + 8x + 14x + 112 = 135

$x^{2}$ + 8x + 14x -23 = 0

$x^{2}$ + 22x -23 = 0

$x^{2}$ + 23x - x - 23 = 0

(x-1)(x+23) = 0

Therefore x = 1

Expanded length = 1+14 = 15

Expanded width = 1+8 = 9

Perimeter = 2(9+15) = 48 units

Learn more about Rectangles: brainly.com/question/25292087

#SPJ1

3 0

1 year ago