Why can't you use the product of powers rule to simplifythis expression? Explain.
3^4.2^8

Answers

Answer 1
Answer:

Answer:

Because the base of this expression is not same.


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Find the slope of the line that contains the points (4,2) and (7,-4)*

Answers

Answer:

-2

Step-by-step explanation:

To find the slope of the line you have to use the equation,

(y2-y1)/(x2-x1)

In this case it is, (-4-2)/7-4)

This simplifies to -2 and this is the slope of the line

Answer:

-8/5

hope this help!

Dr. Smith wants to use only students with "normal" IQ in his experiment. He defines "normal" as anyone who scores in the middle 50% of IQ scores. Using this rule, what will be the lowest IQ score that could be included in the study and what would be the highest IQ score that could be included in the study? You know that the population of IQ scores are normally distributed, have µ = 100, and have σ = 15. (1 point)

Answers

Answer:

Lowest IQ: 89.875

Highest IQ: 110.125

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 100, \sigma = 15.

He defines "normal" as anyone who scores in the middle 50% of IQ scores. Using this rule, what will be the lowest IQ score that could be included in the study and what would be the highest IQ score that could be included in the study?

The middle 50% is the interval from the 25th percentile to the 75th percentile.

Lowest IQ:

This is the measure in the 25th percentile. That is X when Z has a pvalue of 0.25. So it is Z = -0.675

Z = (X - \mu)/(\sigma)

-0.675 = (X - 100)/(15)

X - 100 = 15*(-0.675)

X = 89.875

Highest IQ:

This is the measure in the 75th percentile. That is X when Z has a pvalue of 0.75. So it is Z = 0.675

Z = (X - \mu)/(\sigma)

0.675 = (X - 100)/(15)

X - 100 = 15*(0.675)

X = 110.125

Helen rolls a dice and flips a coin.Calculate the probability that she gets a 4 and a tail.

Answers

Answer:

1/12

Step-by-step explanation:

Probability of 4 = 1/6

Probability tail = 1/2

Multiply them

Answer:

1/12

Step-by-step explanation:

Probability of 4 = 1/6

Probability tail = 1/2

1/6 x 1/2 = 1/12

A student purchases school supplies at a store where the sale tax is 7%. The pre-tax price of the supplies is $48. What is the total cost of the supplies?

Answers

$51.36

1.07 * 48 = 51.36
to add 7% to the pre-price, simply multiply the pre-price by 1.07, do this to automatically get the total. I personally see % as decimals so that helps me figure out where to start.

For his long distance phone service, Bill pays a $3 monthly fee plus 11 cents per minute. Last month, Bill's long distance bill was $16.09. For how many minutes was Bill billed?

Answers

Answer:

1.19 minutes

Step-by-step explanation:

First, subtract the $3 monthly fee:

16.09 - 3

= 13.09

Then, divide this by 11:

13.09/11

= 1.19

So, he was billed for 1.19 minutes

Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. ln(x^2+1)

Answers

Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :-  f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x)  to have a maximum or minimum at given point.

ii)  find first derivative f^(l) (x) and equating zero

iii) solve and find 'x' values

iv) Find second derivative f^(ll)(x) >0 then find the minimum value at x=a

v) Find second derivative f^(ll)(x) <0 then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

step1:- find first derivative f^(l) (x) and equating zero

  f^(l)(x) = (1)/(x^2+1) (d)/(dx)(x^2+1)

f^(l)(x) = (1)/(x^2+1) (2x)  ……………(1)

f^(l)(x) = (1)/(x^2+1) (2x)=0

the point is x=0

step2:-

Again differentiating with respective to 'x', we get

f^(ll)(x)=(x^2+1(2)-2x(2x))/((x^2+1)^2)

on simplification , we get

f^(ll)(x) = (-2x^2+2)/((x^2+1)^2)

put x= 0 we get f^(ll)(0) = (2)/((1)^2)   > 0

f^(ll)(x) >0 then find the minimum value at x=0

Final answer:-

The minimum value of the given function is f(0) = 0