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astraxan [27]
3 years ago
6

Find f(5) if f(t) = ^6 – 3t – 4. A –6 B –2 C 14 D 6

Mathematics
1 answer:
Lera25 [3.4K]3 years ago
8 0
What they said ^ I was just about to answer, but there’s popped up first lol
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Solve 3x-2/4 - 2x-5/3 = 1+x/6
elena-14-01-66 [18.8K]

Answer:

x = 3.8

hope it's helpful ❤❤❤

THANK YOU.

5 0
2 years ago
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Last one, So so sorry!
zvonat [6]
C and D are the same, so its between A and B, which concludes your answer should be B
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Study the Proof: Give the reasons for the missing reason. List the number and then state the reason.
joja [24]
1) Given
2) Adding segments AB and BC together gives AC (Addition postulate, not too sure about the terminology) 
3) Segment Addition postulate 
4) Segment Addition postulate(Knowing that AC=BD, the distance between BC and CD add up to BD, shown in #3. AB and BC add together to give AC) 
5) Subtraction property of Equality 

Hope I helped :) 
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3 years ago
Turn -2x+4y=0 into slope-intercept form
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3 0
3 years ago
The age at which small breed dogs are fully housebroken follows a Normal distribution with mean ms = 6 months and standard devia
atroni [7]

Answer:

This measure is just 0.17 standard deviations from the mean, so we should not be surprised.

Step-by-step explanation:

Normal probability distribution

When the distribution is normal, we use 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 the z-score is lower than -2, or higher than 2.5, the score of X is considered unusual.

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

Let xS – xL represent the sampling distribution.

Mean s 6, means L 4. So

\mu = 6 - 4 = 2

Standard deviation s is 2.5, for L is 1.5. So

\sigma = \sqrt{2.5^2+1.5^2} = 2.915

Should we be surprised if the sample mean housebroken age for the small breed dogs is at least 2.5 months more than the sample mean housebroken age for the large breed dogs? Explain your answer.

We have to find the z-score for X = 2.5. So

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

Z = \frac{2.5 - 2}{2.915}

Z = 0.17

This measure is just 0.17 standard deviations from the mean, so we should not be surprised.

7 0
2 years ago
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