What is the answer ?

What are the roots of the quadratic equation X² + 4X + 4. How many roots are there.

kogti [31]

If we set the equation equal to 0, we can factor it to find its roots:

x² + 4x + 4 = 0
(x + 2)(x + 2) = 0
x = -2

This graph has one root, a double root, at -2. This means that a single point, which must be the vertex of the parabola, touches the x-axis at (-2, 0)

3 0

3 years ago

Describe the location of the point having the following coordinates. positive abscissa, positive ordinate Quadrant I Quadrant II

asambeis [7]

The abscissa is the x coordinate whereas the ordinate is the y coordinate. Both x and y are positive in the first quadrant only.

8 0

3 years ago

The first quartile of a data set is 18, the median is 25, the third quartile is 28,

laila [671]

The outlier would be shown as a dot so the answer is B. 45
The interquartile range is 28 - 18 = 10
A simple rule is that points more than 1.5xIRQ above or below the medians is an outlier and is shown as a point.
28 + 1.5x10 = 43. 45 is above this value.

7 0

1 year ago

HELP PLEASE!!

ANTONII [103]

<span>Volume of the Sphere V = 4/3pi x r^3 So when diameter d = 3 inches => r = 1.5 inches volume of the sphere = 4.188 x r^3 = 4.188 x 1.5^3 = 14.14 in So when diameter d = 8 inches => r = 4 inches volume of the sphere = 4.188 x r^3 = 4.188 x 4^3 = 268.03 in So when diameter d = 9 inches => r = 4.5 inches volume of the sphere = 4.188 x r^3 = 4.188 x 4.5^3 = 381.63 in value for option (a) is 3 x 14.14 = 42.42 inches value for option (b) is 268.03 inches value for option (c) is 381.63 / 2 = 190.815 inches So the correct option would be (b)</span>

5 0

3 years ago

The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged

Temka [501]

Answer:

a) 0.283 or 28.3%

b) 0.130 or 13%

c) 0.4 or 40%

d) 30.6 mm

Step-by-step explanation:

z-score of a single left atrial diameter value of healthy children can be calculated as:

z= $\frac{X-M}{s}$ where

X is the left atrial diameter value we are looking for its z-score

M is the mean left atrial diameter of healthy children (26.7 mm)

s is the standard deviation (4.7 mm)

Then

a) proportion of healthy children who have left atrial diameters less than 24 mm

=P(z<z*) where z* is the z-score of 24 mm

z*= $\frac{24-26.7}{4.7}$ ≈ −0.574

And P(z<−0.574)=0.283

b) proportion of healthy children who have left atrial diameters greater than 32 mm

= P(z>z*) = 1-P(z<z*) where z* is the z-score of 32 mm

z*= $\frac{32-26.7}{4.7}$ ≈ 1.128

1-P(z<1.128)=0.8703=0.130

c) proportion of healthy children have left atrial diameters between 25 and 30 mm

=P(z(25)<z<z(30)) where z(25), z(30) are the z-scores of 25 and 30 mm

z(30)= $\frac{30-26.7}{4.7}$ ≈ 0.702

z(25)= $\frac{25-26.7}{4.7}$ ≈ −0.362

P(z<0.702)=0.7587

P(z<−0.362)=0.3587

Then P(z(25)<z<z(30)) =0.7587 - 0.3587 =0.4

d) to find the value for which only about 20% have a larger left atrial diameter, we assume

P(z>z*)=0.2 or 20% where z* is the z-score of the value we are looking for.

Then P(z<z*)=0.8 and z*=0.84. That is

0.84= $\frac{X-26.7}{4.7}$ solving this equation for X we get X=30.648

5 0

3 years ago