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lozanna [386]
3 years ago
8

Kenya exchanges $200 for euros (€). Suppose the conversion rate is €1 = $1.321. How many euros should Kenya receive? (1 point)

Mathematics
2 answers:
IceJOKER [234]3 years ago
6 0
I believe about €151.40
PtichkaEL [24]3 years ago
5 0
264.20 euros because 200 x 1.321 = 264.20
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Please answer :) I have 7 I need somebody to answer 8 it is a 2 part question!
vazorg [7]

Answer:

question 7 would be option 4 because its saying he has AT LEAST $5, which would mean he DOES have $5 or more. Which is why you would use the "more than or equal to" sign that was used in option 4.

question 8 would be option 1 because its a closed circle that includes the $5 and its going to the right.

Step-by-step explanation:

3 0
2 years ago
A popular soft drink is sold in 2-liter (2000 milliliter) bottles. Because of variation in the filling process, bottles have a m
andrezito [222]

Answer:

a) 0.27% probability that the mean is less than 1995 milliliters.

b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

Step-by-step explanation:

To solve this problem, it is important to understand the normal probability distribution and the central limit theorem

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2000, \sigma = 18

A) If the manufacturer samples 100 bottles, what is the probability that the mean is less than 1995 milliliters?

So n = 100, s = \frac{18}{\sqrt{100}} = 1.8

This probability is the pvalue of Z when X = 1995. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{1995 - 2000}{1.8}

Z = -2.78

Z = -2.78 has a pvalue of 0.0027

So 0.27% probability that the mean is less than 1995 milliliters.

B) What mean overfill or more will occur only 10% of the time for the sample of 100 bottles?

This is the value of X when Z has a pvalue of 1-0.1 = 0.9.

So it is X when Z = 1.28

Z = \frac{X - \mu}{s}

1.28 = \frac{X - 2000}{1.8}

X - 2000 = 1.8*1.28

X = 2002.3

2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.

3 0
2 years ago
Lori has 25 new stamps but she already has 15 old ones how many does she have in total
KonstantinChe [14]
She would have a total of 40 stamps. 
7 0
3 years ago
7x3 how much???????????????????????????????????????
elena55 [62]
7 x 3= 21 you can also use a calculator to help you
8 0
2 years ago
Read 2 more answers
Q5: The probability that event A occurs is 5/7 and the probability that event B occurs is 2/3 . If A and B are independent event
bekas [8.4K]

Answer:

P(A \cap B)

And we can use the following formula:

P(A \cap B)= P(A)* P(B)

And replacing the info we got:

P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476

Step-by-step explanation:

We define two events for this case A and B. And we know the probability for each individual event given by the problem:

p(A) = \frac{5}{7}

p(B) = \frac{2}{3}

And we want to find the probability that A and B both occurs if A and B are independent events, who menas the following conditions:

P(A|B) = P(A)

P(B|A) = P(B)

And for this special case we want to find this probability:

P(A \cap B)

And we can use the following formula:

P(A \cap B)= P(A)* P(B)

And replacing the info we got:

P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476

7 0
2 years ago
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