Solve the system: x + 2 = y ,x2 − 10 = y

Mathematics

1 answer:

kondaur [170]3 years ago

6 0

Answer:

If you wanted the intersecting points there is only one and It's (12, 14)

Hope that could help.

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Simplify the following.<br>4p^3q^2 x-3p^6q^5

skelet666 [1.2K]

Answer:

View picture for answer.

Step-by-step explanation:

4 0

3 years ago

Karin and Hector build two cubes using 1-in cubes. Hector cube has a side length of 12 in. Karin cube has a side length of 14 in

irinina [24]

2 inches, becuase 14in minus 12in is 2.

5 0

3 years ago

Read 2 more answers

Please help me with geometry

spayn [35]

For each sub-problem shown below, I'm using the distance formula to compute the distance between the two points

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For the points (-10,2) and (-2,2), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((-2-(-10))^2+(2-2)^2)

d = sqrt((-2+10)^2+(2-2)^2)

d = sqrt((8)^2+(0)^2)

d = sqrt(64+0)

d = sqrt(64)

d = 8

------------------------------------------------------

For the points (-3,-1) and (5,1), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((5-(-3))^2+(1-(-1))^2)

d = sqrt((5+3)^2+(1+1)^2)

d = sqrt((8)^2+(2)^2)

d = sqrt(64+4)

d = sqrt(68)

------------------------------------------------------

For the points (-3,-2) and (1,3), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((1-(-3))^2+(3-(-2))^2)

d = sqrt((1+3)^2+(3+2)^2)

d = sqrt((4)^2+(5)^2)

d = sqrt(16+25)

d = sqrt(41)

------------------------------------------------------

For the points (-3,-5) and (-2,-4), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((-2-(-3))^2+(-4-(-5))^2)

d = sqrt((-2+3)^2+(-4+5)^2)

d = sqrt((1)^2+(1)^2)

d = sqrt(1+1)

d = sqrt(2)

------------------------------------------------------

For the points (0,0) and (5,5), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((5-0)^2+(5-0)^2)

d = sqrt((5)^2+(5)^2)

d = sqrt(25+25)

d = sqrt(50)

------------------------------------------------------

For the points (1,2) and (1,-10), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((1-1)^2+(-10-2)^2)

d = sqrt((0)^2+(-12)^2)

d = sqrt(0+144)

d = sqrt(144)

d = 12

------------------------------------------------------

For the points (1,2) and (5,2), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((5-1)^2+(2-2)^2)

d = sqrt((4)^2+(0)^2)

d = sqrt(16+0)

d = sqrt(16)

d = 4

------------------------------------------------------

For the points (2,3) and (10,9), the distance is...

d = sqrt((x2-x1)^2+(y2-y1)^2)

d = sqrt((10-2)^2+(9-3)^2)

d = sqrt((8)^2+(6)^2)

d = sqrt(64+36)

d = sqrt(100)

d = 10

------------------------------------------------------

See the attached image for how the answers match up (take note of the letter labels)

3 0

4 years ago

g Q3 [20 points] Suppose that your average heart rate is 72. The average heart rate in your class is 82 and the variance is 25.

Ganezh [65]

Answer:

2.28% of the class has a lower heart rate

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation(which is the square root of the variance) $\sigma$ , 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.

In this problem, we have that:

$\mu = 82, \sigma = \sqrt{25} = 5$