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ahrayia [7]
2 years ago
8

The cost of the five most popular items at an airport gift shop include a $12.99 decorative mug, a $2.99 postcard, a $3.49 keych

ain, a $15.29 t-shirt, and a $7.89 teddy bear. Everleigh is interested in seeing the cost of the items in another currency. The most current currency conversion rates for 1 U.S. dollar are the following:
Country Name of Currency Exchange Rate
China Yuan 6.4651
Denmark Krone 6.3394
Sweden Krona 8.6964

Choose one airport gift shop souvenir and one country. Convert the cost of the chosen souvenir into the currency of that country. Show all necessary mathematical calculations.
Mathematics
1 answer:
Leno4ka [110]2 years ago
3 0

Exchange rate refers value of a currency in another.The cost of the decorative mug is 83.98 China Yuan.

<h3>What is the exchange  rate?</h3>

The term exchange rate has to do with the value of a currency in another currency. Given the fact that international trade is the mainstay of the global economy, it is important that one currency should be converted to another for the purpose of the ease of doing business.

The dollar is often the international currency for trade. Its us always necessary to convert the dollar into the local currency of the country where the business is done using the correct exchange rate.

Now If I want to convert the cost of the decorative mug to China Yuan;

I already know that;

1 US dollar =  6.4651 China Yuan

12.99 US dollar = 12.99 US dollar  * 6.4651 China Yuan/1 US dollar

= 83.98 China Yuan

Hence, the cost of the decorative mug is 83.98 China Yuan.

Learn more about exchange rate:brainly.com/question/22996931

#SPJ1

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Find domain of y=rad20-6x
Evgesh-ka [11]
Domain, on an "even root" context, means, an even root cannot have a negative radicand, since  say for example \bf \sqrt{-25}\ne -5\qquad why?\implies (-5)(-5)=25

so... you end up with an "imaginary value"

so... for the case of 20-6x
"x", the domain, or INPUT
can afford to have any value, so long it doesn't make the radicand negative

to check for that, let us make the expression to 0, and see what is "x" then

\bf 20-6x=0\implies 20=6x\implies \cfrac{20}{6}=x\implies \cfrac{10}{3}=x

now, if "x" is 10/3, let's see \bf \sqrt{20-6\left( \frac{10}{3} \right)}\implies \sqrt{20-20}\implies \sqrt{0}\implies 0

now, 0 is not negative, so the radicand is golden

BUT, if "x" has a value higher than 10/3, the radicand turns negative
for example  \bf x=\frac{11}{3}&#10;\\\\\\&#10;&#10;\sqrt{20-6\left( \frac{11}{3} \right)}\implies \sqrt{20-22}\implies \sqrt{-2}

so.. that's not a good value for "x"

thus, the domain, or values "x" can safely take on, are all real numbers from 10/3 onwards, or to infinity if you wish
5 0
3 years ago
Write 2x=y/3-4 in standard form
Sveta_85 [38]
Look in the attached file to see answer

7 0
3 years ago
Read 2 more answers
Please Answer Quick 65 Points ! !
NeTakaya

The correct answers are:

  • The ordered pair (7, 19) is a solution to the first equation because it makes the first equation true.
  • The ordered pair (7, 19) is not a solution to the system because it makes at least one of the equations false.

Further explanation:

Given equations are:

2x-y = -5

x+3y = 22

We have to check whether the given statements are true or not. In order to find that we have to put the points in the equations

Putting the point in 2x-y = -5

2x - y = -5\\2(7) -19 = -5\\14-19 = -5\\-5 = -5

Putting the point in x+3y=22

7 + 3 (19) = 22\\7 + 57 = 22\\64 \neq 22

The point satisfies the first equation but doesn't satisfy the second. So,

1. The ordered pair (7, 19) is a solution to the first equation because it makes the first equation true.

This statement is true as the point satisfies the first equation

2. The ordered pair (7, 19) is a solution to the second equation because it makes the second equation true.

This Statement is false.

3. The ordered pair (7, 19) is not a solution to the system because it makes at least one of the equations false.

This statement is true.

4. The ordered pair (7, 19) is a solution to the system because it makes both equations true.

This statement is false as the ordered pair doesn't satisfy both equations.

Keywords: Solution of system of equations, linear equations

Learn more about solution of linear equations at:

  • brainly.com/question/13168205
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5 0
3 years ago
1. At a high school, 200 students were asked whether they attended the last soccer game. Of the
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Answer:

girl i don't  know this

Step-by-step explanation:

3 0
3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
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