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likoan [24]
3 years ago
14

When converting from logarithmic to exponential or exponential to logarithmic, which statement is correct?

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
3 0
The anwser would b A

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WILL MARK AS BRAINIEST IF RIGHT
Andre45 [30]

Answer:

10

Step-by-step explanation:

5 0
2 years ago
I don't understand how to do this can someone please help
Margarita [4]
Put x over 1 and start @ -1 on the graph
Then put 2 over 1 and start @ -2 on the graph


Do the same thing with the other graph
3 0
3 years ago
Blood pressure values are often reported to the nearest 5 mmHg (100, 105, 110, etc.). The actual blood pressure values for nine
Paladinen [302]

Answer:

a) 117.4

b) 117.9

c) Option A)  When there is rounding or grouping, the median can be highly sensitive to small change

Step-by-step explanation:

We are given the following data set in the question:

108.6, 117.4, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

n = 9

a) Median of the reported blood pressure values

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.4, 120.0, 121.5, 123.2, 128.4

Median =

\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.4

b) New median of the reported values

Data: 108.6, 117.9, 128.4, 120.0, 103.7, 112.0, 98.3, 121.5, 123.2

Sorted Values: 98.3, 103.7, 108.6, 112.0, 117.9, 120.0, 121.5, 123.2, 128.4

New Median =

\dfrac{9 + 1}{2}^{th}\text{ term} = 5^{th}\text{ term} = 117.9

c) Since median is a position based descriptive statistics, a small change in values can bring a change in the median value as the order of the data may change.

Option A)  When there is rounding or grouping, the median can be highly sensitive to small change

7 0
3 years ago
The radius of the base of a right circular cone is 4 in., and its slant height is 15 in. If the radius and the slant height are
alukav5142 [94]

Answer:

Step-by-step explanation:

radius r = 4 in

slant height L = 15 in

base area = πr² = 16π in²

lateral area = πrL = 60π  in²

surface area = 76π in²

r = 4×6, L = 15×6

base area = (4×6)²π = 16π×36

lateral area = 60π×36

surface area is multiplied by 36

3 0
3 years ago
Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha
Pavel [41]

Answer:

(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14.

(Expand to obtain an equivalent expression for the sphere: x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0)

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}.

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}.

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between \left(x_1, \, y_1, \, z_1\right) and \left(x_2, \, y_2, \, z_2\right) would be:

\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right).

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}.

The equation for a sphere of radius r and center \left(x_0,\, y_0,\, z_0\right) would be:

\left(x - x_0\right)^2 + \left(y - y_0\right)^2 + \left(z - z_0\right)^2 = r^2.

In this case, the equation would be:

\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z - (-6)\right)^2 = \left(\sqrt{56}\right)^2.

Simplify to obtain:

\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z + 6\right)^2 = 56.

Expand the squares and simplify to obtain:

x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0.

8 0
3 years ago
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