All categories

3 years ago

it takes Sarah 50 minutes to walk to her friend's house 1 3/4 miles away. What is her walking pace in miles per hour?

1 answer:

3 0

50 min = 5/6 hours

1 3/4 miles = 7/4 miles

$\frac{\frac{7}{4} }{\frac{5}{6} }$ = $\frac{7}{4} * \frac{6}{5} = \frac{42}{20} = \frac{21}{10} = 2.1$

Therefore, the walking rate is 2.1 miles per hour

You might be interested in

Peter attends 6 lessons each week . A year he attends 52 weeks. But he missed 5 classes how many lessons did he attend during th

VMariaS [17]

You are given the information that he takes six lessons per week. First we will calculate the total lessons he could potentially take per year.

6 • 52 = 312

He could potentially take 312 lessons in one year.

Now that you know this information, you simply subtract the days he missed from that total.

312 - 5 = 307

Your final answer: Peter took 307 lessons during the year.

4 0

3 years ago

(12a + 8) + (4a + 3) + (5 + a) =

PSYCHO15rus [73]

Answer:

17a + 16 =

Step-by-step explanation:

Add all a's

then all numbers w/o letters

you cannot add them together bc it would be like adding apples to carrots (thats what my teacher said)

3 0

3 years ago

In the Ardmore Hotel, 20 percent of the guests (the historical percentage) pay by American Express credit card. (a) What is the

xxMikexx [17]

Answer:

(a) The expected number of guests until the next one pays by American Express credit card is 4.

(b) The probability that the first guest to use an American Express is within the first 10 to checkout is 0.0215.

Step-by-step explanation:

The random variable X can be defined as the number of guests until the next one pays by American Express credit card

The probability that a guest paying by American Express credit card is, p = 0.20.

The random variable X follows a Geometric distribution since it is defined as the number of trials before the first success.

The probability mass function of X is:

$P(X=x)=(1-p)^{x}p;\ x=0,1,2,3...,\ 0$

(a)

The expected value of a Geometric distribution is:

$E(X)=\frac{1-p}{p}$

Compute the expected number of guests until the next one pays by American Express credit card as follows:

$E(X)=\frac{1-p}{p}$

$=\frac{1-0.20}{0.20}$

$=4$

Thus, the expected number of guests until the next one pays by American Express credit card is 4.

(b)

Compute the probability that the first guest to use an American Express is within the first 10 to checkout as follows:

$P(X=10)=(1-0.20)^{10}\times0.20$

$=0.1073741824\times 0.20\\=0.02147483648\\\approx0.0215$

Thus, the probability that the first guest to use an American Express is within the first 10 to checkout is 0.0215.

3 0

3 years ago

Help me solve this problem please

miss Akunina [59]

Answer:

JL = 54

Step-by-step explanation:

JL = JK + KL

JK = KL

8x+11 =14x-1

12 = 6x

x = 2

substitute for x:

8(2)+11 + 14(2)-1 = JL

JL = 16+11+28-1

JL = 54

7 0

3 years ago

Peaches are being sold for 2$ per pound. If x represents the number of pounds of peaches bought and y represents the total cost

ladessa [460]

3x-x+2=4, That should be your answer if that is what your looking for

6 0

3 years ago