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3 years ago

6

Explain how solve 4x + 3 = 7 using the change of base formula log base b of y equals log y over log b. Include the solution for

x in your answer. Round your answer to the nearest thousandth. ANSWER ASAP PLEASE

1 answer:

kvasek [131]3 years ago

8 0

Answer:
x=1
Step-by-step explanation:
Given is an exponential equation

Subtract 3 from both the sides
We get

Taking log

=1

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What is the value of (gof)(4)?

Alinara [238K]

I hope this helps you

7 0

3 years ago

Read 2 more answers

What is 8.07 × 102 in standard form

SOVA2 [1]

Answer:

8070

Step-by-step explanation:

Given:

8.07

x

10

3

8.07

×

10

3

=

8.07

×

1000

=

8070

7 0

3 years ago

Suppose <br> f '(x) = xe^−x^2

iogann1982 [59]

<h3>Answers:</h3>

(a) The function is increasing on the interval (0, infinity)
(b) The function is decreasing on the interval (-infinity, 0)

=======================================================

Explanation:

You should find that the derivative is entirely negative whenever x < 0. This suggests that the function f(x) is decreasing on this interval. So that takes care of part (b).

The interval x < 0 is the same as -infinity < x < 0 which then translates to the interval notation (-infinity, 0)

Similarly, you should find that the derivative is positive when x > 0. So the function is increasing on the interval (0, infinity)

6 0

3 years ago

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

3 years ago

A festival charges $3 for child admission and $5 for adult admision. At the end of the festival they have sold 779 tickets for a

lutik1710 [3]

Answer:

562 child tickets were sold

217 adult tickets were sold

Step-by-step explanation:

A festival charges $3 for children admission and $5 for adult admission

At the end of the festival they have sold a total number of 779 tickets for $2771

Let x represent the child ticket

Let y represent the adult ticket

x + y= 779..............equation 1

3x + 5y= 2771..........equation 2

From equation 1

x + y = 779

x= 779 -y

Substitute 779-y for x in equation 2

3x + 5y= 2771

3(779-y) + 5y= 2771

2337 - 3y + 5y= 2771

2337 +2y= 2771

2y= 2771 -2337

2y = 434

y = 434/2

y = 217

Substitute 217 for y in equation 1

x + y= 779

x + 217= 779

x = 779-217

x= 562

Hence 562 child tickets were sold and 217 adult tickets were sold

6 0

3 years ago