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3 years ago

9

What is the product of 2.5x10-15 and 3.9x1026?

2 answers:

stellarik [79]3 years ago

6 0

Answer:

2.5×3.9×10^(26-15) = 9.75×10^(11)

kozerog [31]3 years ago

3 0

Answer: 9.75 x 10^11

Step-by-step explanation:

Product means multiply so we're solving 2.5x10^-15 x 3.9x10^26.

Since multiplication can be done in any order, we can do this in two stages:

(2.5x3.9) x (10^-15x10^26)

First 2.5 x 3.9 = 9.75

Secondly we have 10^-15 x 10^26. The bases (10) are the same, so all we need to do is add the powers, -15 + 26 = 11. So 10^-15 x 10^26 = 10^11

Finally we put them back together, and have 9.75 x 10^11

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If r(X)=2-x^2 and w(X)=X-2, what is the range of (w*r)(X)

slega [8]

The given function are
r(x) = 2 - x² and w(x) = x - 2
<span>(w*r)(x) can be obtained by multiplying the both function together
</span>
So, <span>(w*r)(x) = w(x) * r(x) = (x-2)*(2-x²)</span>
<span>(w*r)(x) = x (2-x²) - 2(2-x²)</span>
= 2x - x³ - 4 + 2x²

∴ <span>(w*r)(x) = -x³ + 2x² + 2x - 4
</span>

<span>It is a polynomial function with a domain equal to R
</span>
The range of <span>(w*r)(x) can be obtained by graphing the function
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To graph (w*r)(x), we need to make a table between x and (w*r)(x)

See the attached figure which represents the table and the graph of <span>(w*r)(x)
</span>

As shown in the graph the range of <span>(w*r)(x) is (-∞,∞)
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8 0

3 years ago

Read 2 more answers

A four year old is going to spin around with his arms stretched out 100 times. From past experience, his father knows it takes a

Sati [7]

Answer:

$P(X \geq0.55) \leq 0.22$

Step-by-step explanation:

Using central Limit Theorem (CLT), The sum of 100 random variables;

$Y=X_1+X_2+...+X_{100}$ is approximately normally distributed with

Y ~ N (100 × $\frac{1}{3^2}$ ) = N ( 50, $\frac{100}{9}$ )

The approximate probability that it will take this child over 55 seconds to complete spinning can be determined as follows;

N ( 50, $\frac{100}{9}$ )

$P(Y>55) =P(Z>\frac{55-50}{10/3})$

$P(Y>55) =P(Z>1.5)$

$P(Y>55) =\phi (-1.5)$

$P(Y>55) =0.0668$

Using Chebyshev's inequality:

$P(|X-\mu\geq K)\leq \frac{\sigma^2}{K^2}$

Let assume that X has a symmetric distribution:

Then:

$2P(X-\mu\geq K)\leq) \frac{\sigma^2}{K^2}$

$2P(X \geq \mu+K)\leq) \frac{\sigma^2}{K^2}$

$2P(X\geq0.5+0.05)\leq \frac{\frac{1}{\frac{3^2}{100} } }{0.05^2}$ where: ( $\sigma^2 = \frac{1}{3^2/100}$ )

$P(X \geq0.55) \leq 0.22$

6 0

3 years ago

PLZ HELP ME IM SO CONFUSED

r-ruslan [8.4K]

A = (pi * r^2)/2
A = (3.1416 * (14.3/2)^2)/2
A = 80.3 ft^2

6 0

3 years ago

4) linear equations <br><br> 9+4x=-15

Sunny_sXe [5.5K]

Answer:

X=-6

Step-by-step explanation:

Subtract 9 on both sides

Add -9 and -15 You get -24

Divide -24 by 4 you get x=-6

8 0

3 years ago

The function m(x) is the function f(x) =3(x-5)^2+5 transformation by horizontal shift 4 units left and vertical shift of 3 units

maxonik [38]

3(x - 5)^2 + 5
vertex: (5, 5)
x = 5 - 4 = 1
y = 5 + 3 = 8
new vertex: (1, 8)
3(x - 1)^2 + 8

Answer:
(second option) 3(x - 1)^2 + 8

8 0

2 years ago