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katrin [286]
2 years ago
5

Find the value of x. (6x + 1) (9x - 23)

Mathematics
1 answer:
shtirl [24]2 years ago
3 0

Answer:

Step-by-step explanation:

(6x+1)(9x-23)= 6x(9x-23)+1(9x-23)

= 54x²-138+9x-23

= 54x²+9x-161

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How many solutions can have two intersecting lines have?<br> A. 3<br> B. 0 <br> C. 1
Vladimir [108]

two intersecting lines can only have 1 solution.

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2.) Use the Slope Intercept Form of a line to find the equation of the line from point C to point D.
mezya [45]

First we need slope

  • C=(0,0)
  • D(7,12)

\\ \sf\longmapsto m=\dfrac{12-0}{7-0}

\\ \sf\longmapsto m=\dfrac{12}{7}

Put D co-ordinates on y=mx+b

\\ \sf\longmapsto 12=\dfrac{12}{7}(7)+b

\\ \sf\longmapsto 12=12+b

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slope intercept form.

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2 years ago
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Find the median if this set of numbers: 7, 6, 19, 12, 16, 19, 19, 6 ​
tankabanditka [31]

Answer:

20

Step-by-step explanation:

Order the numbers from smallest to largest, then cross one number off until you have one or two numbers left.

If there is an EVEN number of numbers(like this one): there will be two numbers left, therefore you add those two numbers together, then divide by two.

If there is  an ODD number of numbers: there will be one number left, therefore that is your answer!

Hope this helps!

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6 0
3 years ago
a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was
Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

<em>Let </em>\bar X<em> = sample average time spent studying</em>

The z-score probability distribution for sample mean is given by;

          Z = \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }  ~ N(0,1)

where, \mu = population mean hours spent studying = 25 hours

            \sigma = standard deviation = 15 hours

            n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < \bar X < 30 hours)

    P(29 hours < \bar X < 30 hours) = P(\bar X < 30 hours) - P(\bar X \leq 29 hours)

      

    P(\bar X < 30 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } < \frac{ 30-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z < 2) = 0.97725

    P(\bar X \leq 29 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } \leq \frac{ 29-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z \leq 1.60) = 0.94520

                                                                    

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.</em>

Therefore, P(29 hours < \bar X < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0
2 years ago
Reduced ratio 27 to 9
Sidana [21]
3:1 is your answer :)
3 0
3 years ago
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