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Jet001 [13]
4 years ago
5

Evaluate x/y for x= 2/3 and y= 5/6

Mathematics
2 answers:
Fiesta28 [93]4 years ago
8 0

4/5 or 0.8

Step-by-step explanation:

x = ⅔

y = 5/6

\frac{x}{y}  =  \\  \frac{2}{3}   \div  \frac{5}{6}

Change the sign and flip the value on the left

\frac{2}{3}  \times  \frac{6}{5}  \\  =  \frac{12}{15}

Simplify

=  \frac{4}{5}

kykrilka [37]4 years ago
7 0

Answer:

4/5

Step-by-step explanation:

by substituting the value given in the question u get the answer 4/5

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Which transformations could be performed to show that
andre [41]

Answer: a 180 degree rotation about the origin, then a dilation by a scale factor of 1/3

Step-by-step explanation:

3 0
3 years ago
The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need 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.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
3 years ago
I need this really quick please.
Anestetic [448]
The correct answer is 7x-8
7 0
3 years ago
Read 2 more answers
Katherine drove 13 miles in 1/5 hour. On average, how fast did she drive, in miles<br> per hour?
skelet666 [1.2K]

Answer:

65 miles per hour

Step-by-step explanation:

r = d/t

1 : 5 = 0.2

r = 13 : 0.2 = 65

4 0
3 years ago
Please help.
34kurt
The number would be greater. if you think about normal multiplication, when you multiply a number by 1 the number stays the same, but when you multiple the number by less then one, i.e. a fraction you get a fraction of what you had. while a number greater then one, i.e. 2, the number increases. so 1.23 is greater then one so it would increase
6 0
3 years ago
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