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serious [3.7K]
2 years ago
5

Can you guys let me know if I'm right please

Mathematics
1 answer:
MAXImum [283]2 years ago
3 0

Answer:

No there are problems. Please correct the errors

1 \frac{3}{20} = \frac{23}{20} = 23:20

25\% = \frac{1}{4} = 0.25

Step-by-step explanation:

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If x=3+root8 and y =3- root 8 find 1/x^2+1/y^2
KonstantinChe [14]

Answer:

34

Step-by-step explanation:

x=3+√8

y =3-√8

now,

1/x^2+1/y^2

=1/(3+√8)² + 1/(3-√8)²

= [(3-√8)²+(3+√8)²] / (3+√8)²(3-√8)²            [L.C.M  = (3+√8)²(3-√8)² ]

=[(3-√8+3+√8)²-2(3-√8)(3+√8) ] / [(3+√8)(3-√8)]²  

=[6²-2.(3²-√8² )] / (3²-√8²)²                 [ a²+ b²=(a+b)²-2ab]

=[36-2(9-8) ]/ (9-8)²

=[36-2.1] / 1²

=34

3 0
3 years ago
The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of m
UkoKoshka [18]

Answer:

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of \bar X Round your answers to two decimal places.

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

4 0
3 years ago
You guys, I seriously need help with this! I have been working on this for like 5 hours and I still can't understand it! Please
Alina [70]

Step-by-step explanation:

Q1. E

Q2. A

Q3. H

sorry if I'm wrong anyways

Q4. C

4 0
3 years ago
What is the domain of the function f(x) = x + 2 ? a.    all real numbers greater         than -2 b.    all real numbers greater
dsp73
The domain of the function is represented by option C. All real numbers.
3 0
3 years ago
3/4 times the amount of 20
klio [65]

Answer:

= { \rm{ \frac{3}{4}  \times 20}} \\  \\  = { \rm{ \frac{60}{4} }} \\  \\  = { \boxed{ \tt{ \:  \: 15 \:  \: }}}

8 0
2 years ago
Read 2 more answers
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