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3 years ago

FIND THE PRODUCTS OR QUOTIENYS IN THE EXPONENTIAL FORMS BY USING LAWS OF INDICES.

Mathematics

1 answer:

hram777 [196]3 years ago

8 0

Answer:

(5x)^-5.

Step-by-step explanation:

(5x)^3 * (25x^2)^2 * (5x)^-12

= (5x)^3 * (5x)^4 + (5x)^-12

= (5x)^(3+4-12)

= (5x)^-5.

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Which statements are true regarding f(x)=10/x+5

belka [17]

Answer:

1. The domain of f(x) is (–∞, –5) U (–5, ∞).

4. The y-intercept is (0, 2).

5. There is a vertical asymptote at x = –5.

6. The end behavior is x → –∞, f(x) → 0 and x → ∞, f(x) → 0.

Step-by-step explanation:

I got it correct on e2020

5 0

3 years ago

Read 2 more answers

Divide. Round your answer to the nearest whole number.<br>91,543 = 5 =

aliina [53]

Answer:

i assume you mean divide so its 18308

Step-by-step explanation:

3 0

3 years ago

(5s-6)(3s+2) please just give me the answer I have to turn it in today

alexandr1967 [171]

Answer:

The answer is -2

Step-by-step explanation:

you subtract common numbers, so:

5s - 3s = 2s

Then,

-6 + 2 = -4

You would then get

2s - 4

divide -4 by 2 which is -2

and your answer is s = -2

5 0

4 years ago

An individual repeatedly attempts to pass a driving test. Suppose that the probability of passing the test with each attempt is

vladimir1956 [14]

Answer:

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

b) $P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

c) $P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

Part b

We want this probability:

$P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

We find the individual probabilities like this:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

Part c

For this case we want this probability:

$P(X \geq 5)$

And we can use the complement rule like this:

$P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

3 0

3 years ago

B) y = log2 (4" - 4)<br> find the inverse of this equation

GuDViN [60]

Answer:

$y = \frac{1}{4} \times {2}^{x} - 1$

Step-by-step explanation:

Assuming the given logarithmic equation is

$y = \log_{2}( {4}{x } - 4)$

We interchange x and y to get:

$x= \log_{2}( {4}{y} - 4)$

We solve for y now:

${2}^{x} = {4}{y} - 4$

We add 4 to both sides to get;

${2}^{x} + 4 = 4y$

Divide through by 4:

$y = \frac{1}{4} \times {2}^{x} - 1$

4 0

3 years ago

FIND THE PRODUCTS OR QUOTIENYS IN THE EXPONENTIAL FORMS BY USING LAWS OF INDICES.​

FIND THE PRODUCTS OR QUOTIENYS IN THE EXPONENTIAL FORMS BY USING LAWS OF INDICES.