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allochka39001 [22]
2 years ago
8

Myra asked Annabelle to buy her some perfume while she was in France. Annabelle paid 200 euros for it. Myra wants to pay Annabel

le back in U.S. dollars. If the foreign exchange rate between the U.S. dollar and the euro is 1:0.7, how much should Myra pay Annabelle?
A.) $140
B.) $240
C.) 286
Mathematics
1 answer:
serious [3.7K]2 years ago
6 0

Answer:

286

Step-by-step explanation:

Given : Myra asked Annabelle to buy her some perfume while she was in France. Annabelle paid 200 euros for it. Myra wants to pay Annabelle back in U.S. dollars.

To Find:  If the foreign exchange rate between the U.S. dollar and the euro is 1:0.7, how much should Myra pay Annabelle?

Solution:

Since Annabelle paid 200 euros

And Myra wants to pay Annabelle back in U.S. dollars.

Let Myra paid amount to Annabelle in U.S. dollar be x

We are also given that  the U.S. dollar and the euro is 1:0.7

⇒\frac{1}{0.7}=\frac{x}{200}

⇒\frac{1*200}{0.7}=x

⇒\frac{200}{0.7}=x

⇒285.7=x

⇒285.7≈286

Thus Myra should pay $286 to Annabelle



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How many more barrels of gasoline then diesel were produced last year gasoline 34% diesel 29% total crude oil -60 million barrel
MArishka [77]

Answer: 3 million barrels


Step-by-step explanation:

Given :- Total crude oil= 60 million barrels

Quantity of gasoline = 34% of total crude oil

=0.34\times60\text{ million barrels}\\=20.4\text{ million barrels}

Quantity of diesel = 29% of total crude oil

=0.29\times60\text{ million barrels}\\=17.4\text{ million barrels}

Clearly, Quantity of gasoline is more than Quantity of diesel

difference = Quantity of gasoline-Quantity of diesel

20.4\text{ million barrels}-17.4\text{ million barrels}\\=3\text{ million barrels}

Therefore, The production of gasoline is 3 million barrels more than diesel last year.



3 0
2 years ago
Read 2 more answers
Express f(x) = |x-2| +|x+2| in the non-modulus form. Hence, sketch the graph of f.
alexgriva [62]
Recall that

|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x

There are three cases to consider:

(1) When x+2, we have |x+2|=-(x+2) and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)-(x+2)=-2x-4

(2) When x+2\ge0 and x-2, we get |x+2|=x+2 and |x-2|=-(x-2), so

|x-2|+|x+2|=-(x-2)+(x+2)=4

(3) When x-2\ge0, we have |x+2|=x+2 and |x-2|=x-2, so

|x-2|+|x+2|=(x-2)+(x+2)=2x

So

|x-2|+|x+2|=\begin{cases}-2x-4&\text{if }x
4 0
2 years ago
A shoe manufacturer compared material A and material B for the soles of shoes. Twelve volunteers each got two shoes. The left wa
sveta [45]

Answer:

a) Are dependent since we are mesuring at the same individuals but on different times and with a different method

b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

c) p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation  

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

(a) Are the two samples paired or independent? Explain your answer.

Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference d_i=A_i-B_i and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

qqnorm(d)

And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

The system of hypothesis for this case are:

Null hypothesis: \mu_A- \mu_B = 0

Alternative hypothesis: \mu_A -\mu_B \neq 0

The second step is calculate the mean difference  

\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{80}{12}=6.67

The third step would be calculate the standard deviation for the differences, and we got:

s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =23.849

The 4 step is calculate the statistic given by :

t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{6.67 -0}{\frac{23.849}{\sqrt{12}}}=0.969

The next step is calculate the degrees of freedom given by:

df=n-1=12-1=11

Now we can calculate the p value, since we have a two tailed test the p value is given by:

p_v =P(t_{(12)}>0.969) =0.353

So the p value is higher than the significance level given, so then we can conclude that we FAIL to reject the null hypothesis. So we can conclude that the mean differences is NOT significantly different from 0 .

4 0
2 years ago
Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum
dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

Thus it 30.671% probable that 0, 1, or 2 students received accommodation.

e.

The expected value from d) is .6

The average time is [.6(4.5) + 19.2(3)]/30 = 2.01 hours

8 0
2 years ago
Playing tennis burns energy at a rate of about 25 kJ/min. Cycling burns energy at about 35kJ/min. Hans exercised by playing tenn
grigory [225]
To answer this, you need to make a systems of equations.

25x + 35y = 1450
x + y = 50 ----> y = 50 - x

25x + 35(50-x) = 1450
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-10x + 1750 = 1450
- 1750 -1750

-10x = -300
x = 30


He played tennis for 30min


8 0
2 years ago
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