A rectangle has an area of 72 in². The length and the width of the rectangle are changed by a scale factor of 3.5.

Problem We can calculate EEE, the amount of euros that has the same value as DDD U.S. Dollars, using the equation E=\dfrac{17}{2

Tanzania [10]

Complete Question

We can calculate EEE, the amount of euros that has the same value as DDD U.S. Dollars, using the equation

E= 17/20 D

a) How many euros have the same value as 1 U.S. Dollar? euros

b) How many U.S. Dollars have the same value as 1 euro? dollars

Answer:

a) 0.85 euros

b) 1.18 dollars

Step-by-step explanation:

The equation is given as:

E= 17/20 D

a) How many euros have the same value as 1 U.S. Dollar? euros

E = 17/20D

D = 1 dollar

Hence:

E = 17/20 × 1

E = 0.85 euros

b) How many U.S. Dollars have the same value as 1 euro? dollars

E = 17/20D

E = 1 euro

1 = 17/20D

D = 1 ÷ 17/20

D = 1 × 20/17

D = 1.1764705882 dollars

D = 1.18 dollars

3 0

3 years ago

Two weeks in a row, the golf course hosts a group of golfers. The second week had 10 more golfers than the first week. Use the d

Trava [24]

Answer:

Option A.

Step-by-step explanation:

From the first dot plot the given data set for week 1 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69), (70, 70, 72, 72, 72), (75, 75, 76, 78, 78), (80, 80, 80, 89, 89)

$Range=Maximum-Minimum=89-62=27$

$Median=\frac{72+75}{2}=73.5$

$Mean=\frac{\sum x}{n}=\frac{1491}{20}=74.55$

From the second dot plot the given data set for week 2 is

62, 68, 68, 68, 69, 70, 70, 72, 72, 72, 75, 75, 76, 78, 78, 80, 80, 80, 85, 86, 86, 86, 88, 88, 88, 88, 89, 89, 89, 89

Divide the data in 4 equal parts.

(62, 68, 68, 68, 69, 70, 70), 72, (72, 72, 75, 75, 76, 78, 78), (80, 80, 80, 85, 86, 86, 86), 88, (88, 88, 88, 89, 89, 89, 89)

$Range=Maximum-Minimum=89-62=27$

$Median=\frac{78+80}{2}=79$

$Mean=\frac{\sum x}{n}=\frac{2364}{30}=78.8$

The range for Week 1 is equal to the range for Week 2.

The median for Week 2 is more than the median for Week 1.

The mean for Week 2 is more than the mean for Week 1.

Therefore, the correct option is A.

6 0

3 years ago

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

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^{-\mu}*\mu^{x}}{(x)!}
$

In which

x is the number of sucesses

$
e = 2.71828$ is the Euler number

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

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

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

In which

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

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

8 0

3 years ago

Express the ratio as a fraction in the simplest form:<br><br> 32 pennies to 300 pennies

jenyasd209 [6]

Answer:

8/75

Step-by-step explanation:

Ratio:

32:300

Fraction:

32/300

Simplest form(divide the numerator and the denominator by 4)

Answer: 8/75

4 0

3 years ago

Read 2 more answers

I’ll put the data on another question please help!!!

bogdanovich [222]

Sorry no cluegvxfbhefgvwsgfregvrgvegbve n eberg

3 0

3 years ago