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9966 [12]
3 years ago
15

If you were to solve the following system by substitution, what would be the best variable to solve for and from what equation?2

x+6y=9
3x-12y=15
Mathematics
1 answer:
aivan3 [116]3 years ago
8 0

The second equation because they all have a common factor of 3. You can get x by itself. However, with the first one, you'll end up with fractions instead.

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The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a
AlekseyPX

Answer:

The time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

Step-by-step explanation:

We are given that the time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a standard deviation of 40 seconds.

The fitness association wants to recognize the fastest 10% of the boys with certificates of recognition.

<u><em>Let X = time for this event for boys in secondary school</em></u>

SO, X ~ Normal(\mu=460,\sigma^{2} =40^{2})

The z-score probability distribution for normal distribution is given by;

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

where, \mu = mean time = 460 seconds

            \sigma = standard deviation = 40 seconds

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.

<u>Now, it is given that the fitness association wants to recognize the fastest 10% of the boys with certificates of recognition, which means;</u>

       P(X > x) = 0.10   {where x is the required time which boy need to beat}

       P( \frac{X-\mu}{\sigma} > \frac{x-460}{40} ) = 0.10

        P(Z > \frac{x-460}{40} ) = 0.10

<em>So, the critical value of x in the z table which represents the top 10% of the area is given as 1.2816, that is;</em>

<em>                   </em>      \frac{x-460}{40} =1.2816

                         {x-460}{} =1.2816\times 40

<em>                           </em>x = 460 + 51.264 = <u>511.264 seconds</u>

Hence, the time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

8 0
3 years ago
Lines a and b are parallel Line cis perpendicular to both line a and line b. Which
prohojiy [21]

Answer:

  see below

Step-by-step explanation:

The slopes of parallel lines are the same. The slopes of perpendicular lines are negative reciprocals of each other, hence their product is -1.

___

For the most part, the concept of adding slopes of lines does not relate to parallel or perpendicular lines in any way.

6 0
3 years ago
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Which term has the definition "The value that describes the location of a digit within a number, such
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Answer:

place value is the answer

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The height of a right circular cylinder is 1.5 times the radius of the base. What is the ratio of the total surface area to the
Naily [24]

Let r represent the radius of cylinder.

We have been given that the height of a right circular cylinder is 1.5 times the radius of the base. So the height of the cylinder would be 1.5r.

We will use lateral surface area of pyramid to solve our given problem.

LSA=2\pi r h, where,

LSA = Lateral surface area of pyramid,

r = Radius,

h = height.

Upon substituting our given values in above formula, we will get:

LSA=2\pi r\cdot (1.5)r  

Now we will find the total surface area of cylinder.

TSA=2\pi r(r+h)

TSA=2\pi r(r+1.5r)

TSA=2\pi r(2.5r)

\frac{TSA}{LSA}=\frac{2\pi r(2.5r)}{2\pi r(1.5r)}

\frac{TSA}{LSA}=\frac{2.5r}{1.5r}

\frac{TSA}{LSA}=\frac{25}{15}

\frac{TSA}{LSA}=\frac{5}{3}

Therefore, the ratio of total surface area to lateral surface area is 5:3.

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If two or more polygons are congruent, which statement must be true about the polygons? Read all of the choices before deciding.
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Pairs of corresponding angles have the same measure.
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