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tatiyna
3 years ago
15

What is the value of X

Mathematics
2 answers:
enyata [817]3 years ago
5 0

If x=6 and y=3 is 9_______________________________________

satela [25.4K]3 years ago
5 0

Answer:

x = 2

Step-by-step explanation:

The product of the external part times the entire secant of one secant is equal to the external part times the entire secant of the other secant, that is

(x + 4)(x + 5) = (x + 1)(x + 12) ← distribute factors on both sides

x² + 9x + 20 = x² + 13x + 12 ( subtract x² from both sides )

9x + 20 = 13x + 12 ( subtract 13x from both sides )

- 4x + 20 = 12 ( subtract 20 from both sides )

- 4x = - 8 ( divide both sides by - 4 )

x = 2

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In a school with 200 students, 45% are males. how many males are there.? please help me​
LUCKY_DIMON [66]

Answer:

<u><em>90 males.</em></u>

Step-by-step explanation:

To find the answer we will <u><em>divide 45% by 100</em></u>, and we get <u><em>0.45</em></u>.  Now <u><em>multiply 0.45 by 200</em></u> and you get <u><em>90</em></u>.  Therefore, <u><em>90 males is the answer. </em></u>

7 0
3 years ago
Write a rule describing the translation below
CaHeK987 [17]

Answer:

d=e

Step-by-step explanation:

3 0
3 years ago
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
This will be so helpful right now thank
RoseWind [281]

Answer:

55 ft.

Step-by-step explanation:

First find the area of Lot X

67 • 70 = 4,690

Subtract the are of Lot X from the total area

8,375 - 4,690 = 3,685

3,685 is the area of Lot Y

Now divide 3,685 by the length of Lot Y : 67

3,685 ÷ 67 = 55

The width of Lot Y is 55 ft.

Hope this helps!

5 0
3 years ago
What are the best approximations of the solutions to this system?
Scilla [17]

(-1.2,-2.0) and (1.9,2.2) are the best approximations of the solutions to this system.

Option B

<u>Step-by-step explanation:</u>

Here, we have a graph of two functions from which we need to find the approximate value of common solutions. Let's find this:

First look at where we have intersection points, In first quadrant & in third quadrant.

<u>At first quadrant:</u>

Draw perpendicular lines from x-axis & y-axis from this point . After doing this we can clearly see that the perpendicular lines cut x-axis at x=1.9 and y-axis at y=2.2. So, one point is (1.9,2.2)

<u>At Third quadrant:</u>

Draw perpendicular lines from x-axis & y-axis from this point. After doing this we can clearly see that the perpendicular lines cut x-axis at x=-1.2 and y-axis at y=  -2.0. So, other point is (-1.2,-2.0).

5 0
3 years ago
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