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antoniya [11.8K]
3 years ago
6

3 + 3(k + 3) = 6(k - 2) +9​

Mathematics
1 answer:
ANTONII [103]3 years ago
7 0

Step-by-step explanation:

3 + 3(k + 3) = 6(k - 2) + 9

3 + 3k + 9 = 6k - 12 + 9

12 + 12 - 9 = 6k - 3k

15 = 3k

k = 15/3

k = 5

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Which best explains whether or not all isosceles triangles are similar?
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The third statement is correct.
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6 0
3 years ago
In the quadratic function y = ax^2 + bx + c, the ___________ affects the width of the parabola.
Gre4nikov [31]

Answer:

the answer is the a affects the width

Step-by-step explanation:

the answer is a not the choice A, the third one

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6 0
3 years ago
calculate the means and the standard deviation of the following set of data 4.578g, 4.581g, 4.572g, 4.573g, 4.601g, 4.577g. stat
AfilCa [17]

Answer:

68% Confidence interval  = [4.5752, 4.5848]

95% Confidence interval  = [4.5688, 4.5918]

Step-by-step explanation:

Sample mean (X) = 4.580

Sample Standard Deviation (S) = 0.01065

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T_{(5)} for alpha/2 0.975 = 2.5706

68% Confidence interval = [x-T_{(5)}\frac{S}{\sqrt{n}}, x+T_{(5)]\frac{S}{\sqrt{n}}] = [4.5752, 4.5848]

95% Confidence interval = [x-T_{(5)}\frac{S}{\sqrt{n}}, x+T_{(5)}\frac{S}{\sqrt{n}}] = [4.5688, 4.5918]

7 0
3 years ago
Calls to a customer service center last on average 2.8 minutes with a standard deviation of 1.4 minutes. An operator in the call
Natali5045456 [20]

Answer:

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

n-values of normal variable:

Suppose we have n values from a normally distributed variable. The mean of the sum of all the instances is M = n\mu and the standard deviation is s = \sigma\sqrt{n}

Calls to a customer service center last on average 2.8 minutes.

This means that \mu = 2.8

75 calls each day.

This means that n = 75

What is the expected total amount of time in minutes the operator will spend on the calls each day

This is M, so:

M = n\mu = 75*2.8 = 210

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

6 0
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If you click on this question i will give you a free website for turtoing <br> in comments
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