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2 years ago

8

Ava wants to figure out the average speed she is driving. She starts checking her car’s clock at mile marker 0. It takes her 4 m

inutes to reach mile marker 3. When she reaches mile marker 6, she notes that 8 minutes total have passed since mile marker 0.
What is Ava’s speed in miles per hour?

2 answers:

Novosadov [1.4K]2 years ago

6 0

Answer:

45 mph

Step-by-step explanation:

Find the unit rate (i. e., the speed):

6 mi 60 min

--------- * ------------- = (3/4)(60) mph = 45 mph

8 min 1 hr

iren [92.7K]2 years ago

6 0

Answer:

0.75 miles per minute

Step-by-step explanation:

You might be interested in

The number of hurricanes hitting the coast of Florida annually has a Poisson distribution with a mean of 0.8. Answer the followi

slava [35]

Answer:

a) $P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]$

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

b) $P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

Step-by-step explanation:

Let X the random variable that represent the number of hurricanes hitting the coast of Florida annualle. We know that $X \sim Poisson(\lambda=0.8)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda=0.8$

$E(X)=\mu =\lambda=0.8$

Part a

For this case we want this probability: $P(X>2)$

And for this case we can use the complement rule like this:

$P(X>2)= 1-P(X \leq 2) = 1-[P(X=0)+P(X=1)+P(X=2)]$

And we can find the individual probabilities like this:

$P(X=0) = \frac{e^{-0.8} 0.8^0}{0!}= 0.4493$

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

$P(X=2) = \frac{e^{-0.8} 0.8^2}{2!}= 0.1438$

And replacing we got:

$P(X>2)= 1-P(X \leq 2) = 1-[0.4493+0.3595+0.1438]=0.0474$

Part b

Using the probability mass function we have:

$P(X=1) = \frac{e^{-0.8} 0.8^1}{1!}= 0.3595$

3 0

2 years ago

The area of the rectangle is 64.8 square centimeters. What is the perimeter of the rectangle? One group of ten tenths and one gr

9966 [12]

Answer:

(a) $Perimeter = 32.2\ cm$

(b) 100

Step-by-step explanation:

Solving (a):

Given

Shape: Rectangle

$Area = 64.8$

Required

Calculate the perimeter.

Area is calculated as:

$Area = L * W$

Where

$L = Length$ and $W = Width$

Substitute 64.8 for Area

$64.8 = L * W$

Make L the subject:

$L = \frac{64.8}{W}$

Perimeter is calculated as:

$P = 2 * (L + W)$

Substitute 64.8/W for L

$P = 2 * (\frac{64.8}{W} + W)$

$P = \frac{129.6}{W} + 2W$

To solve further, we take the derivative of P and set it to 0, afterwards.

$dP = -\frac{129.6}{W^2} + 2$

Set to 0

$0 = -\frac{129.6}{W^2} + 2$

Collect Like Terms

$\frac{129.6}{W^2} = 2$

Cross Multiply:

$2W^2= 129.6$

Divide through by 2

$W^2 = 64.8$

Take square roots

$W = \sqrt{64.8$

$W = 8.05$

Recall that:

$L = \frac{64.8}{W}$

So:

$L= \frac{64.8}{8.05}\\$

$L= 8.05$

The perimeter is:

$Perimeter = 2 * (8.05 + 8.05)$

$Perimeter = 2 * (16.10)$

$Perimeter = 32.2\ cm$

Solving (b):

Given

((1 Group of 10 tenths) and (1 group of 8 tenths))/(6 groups of 3 tenths)

Required

Solve

$Tenths = \frac{1}{10}$

So, the expression becomes:

((1 Group of $10 * \frac{1}{10}$ ) and (1 group of $8 * \frac{1}{10}$ ))/(6 groups of $3 * \frac{1}{10}$ )

This gives:

((1 Group of $\frac{10}{10}$ ) and (1 group of $\frac{8}{10}$ ))/(6 groups of $\frac{3}{10}$ )

Group means product, so the expression becomes:

$\frac{(1 * \frac{10}{10} \ and\ 1 * \frac{8}{10})}{6 * \frac{3}{10}}$

And, as used here means addition

$\frac{(1 * \frac{10}{10} + 1 * \frac{8}{10})}{6 * \frac{3}{10}}$

Simplify:

$\frac{(1 * 1 + 1 * 0.80)}{6 *0.30}$

$\frac{(1 + 0.80)}{1.80}$

$\frac{1.80}{1.80}$

$= 1.00$

7 0

2 years ago

there are 64 singers in the choir. the tenors and sopranos are in separate row . there are 8 singers in each row. there are 4 ro

lilavasa [31]

4 since 64/8=8 there are 8 rows and if 4 are filled with tenors the rest must be filled with sopranos

8 0

2 years ago

Two movie tickets and 3 snacks are $24. Three movies tickets and 4 snacks are $35. How much is a movie ticket and how much is a

Vsevolod [243]

Answer: A movie ticket is $9 while a snack is $2

Step-by-step explanation: We shall let a movie ticket be m while a snack is s. So, from the clues given, if two movie tickets and three snacks cost $24, we can write it as the following expression;

2m + 3s = 24

Also if three movie tickets and four snacks cost $35, we can as well write another expression as follows;

3m + 4s = 35.

Now we have a pair of simultaneous equations which are

2m + 3s = 24 ----------(1)

3m + 4s = 35 ----------(2)

We shall solve this by using the elimination method, since none of the variables has a coefficient of 1. We'll start by multiplying equation (1) by 3 and multiplying equation (2) by 2 (so as to eliminate the m variable)

2m + 3s = 24 -------- x3

3m + 4s = 35 ---------x2

We now arrive at the following

6m + 9s = 72--------(3)

6m + 8s = 70--------(4)

Subtract equation (4) from equation (3) and we arrive at

s = 2

Having determined that s equals 2 we can now substitute for the value of a into equation (1)

2m + 3s = 24

2m + 3(2) = 24

2m + 6 = 24

Subtract 6 from both sides of the equation

2m + 6 - 6 = 24 - 6

2m = 18

Divide both sides of the equation by 2

m= 9

Therefore one movie ticket costs $9 while one snack costs $2

6 0

2 years ago

The volume of air inside a rubber ball with radius r can be found using the function V(r) = four-thirds pi r cubed. What does V

aleksklad [387]

Answer:

The answer is explained below

Step-by-step explanation:

Given that The volume of air inside a rubber ball with radius r can be found using the function V(r) = $\frac{4}{3}\pi r^3$ , this means that the volume of the air inside the rubber ball is a function of the radius of the rubber ball, that is as the radius of the rubber ball changes, also the volume of the ball changes.

As seen from the function, the radius is directly proportional to the volume of the ball, if the radius increases, the volume also increases.

$V(\frac{5}{7} )$ is equal to the volume of the ball when the radius of the ball is $\frac{5}{7}$ . Therefore:

$V(\frac{5}{7} )=\frac{4}{3} \pi *(\frac{5}{7} )^3=1.53\ unit^3$

3 0

2 years ago