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inessss [21]
4 years ago
5

Where does the function f have a zero? (more than one can apply)

Mathematics
1 answer:
alekssr [168]4 years ago
3 0
Answers: 
Choice A(-2,-1)
Choice B(-1,0)
Choice E(2,3)

--------------------------------

Explanation:

The idea is to look for sign changes in f(x). Recall that y = f(x). If the sign of y changes, then y = 0 must be somewhere in the interval if we are to have a continuous function.

The sign changes happen when we go from...
y = -9 to y = +8 (row 2 to row 3)
y = +8 to y = -5 (row 3 to row 4)
y = -1 to y = +196 (bottom two rows)

Those three sign changes correspond to the following interval notations for x
(-2,-1)
(0,1)
(2,3)
in that exact order

Meaning for instance, that a root must be somewhere in the interval -2 < x < -1 since we have a sign change when we transition from f(-2) to f(-1)

So that's why the answers are choices A, B, & E

Something like the interval notation (1,2) will not work because y = -18 does not change sign to y = -1, which is why D is not an answer. Choice C is similar. 

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yan [13]
The starting percent is 100%. After one hour, bacteria tripled, so it is 300%.
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3 years ago
Question 8 of 25<br> Which of the following is the quotient of the rational expressions shown here?
Alla [95]

Answer:

  B.  (5x+5)/(x²-2x)

Step-by-step explanation:

As with numerical fractions, division by a rational expression is equivalent to multiplication by its reciprocal.

<h3>As a multiplication problem</h3>

  \dfrac{5}{x-2}\div\dfrac{x}{x+1}=\dfrac{5}{x-2}\times\dfrac{x+1}{x}

<h3>Product of rational expressions</h3>

As with numerical fractions, the product is the product of numerators, divided by the product of denominators.

  =\dfrac{5(x+1)}{x(x-2)}=\boxed{\dfrac{5x+5}{x^2-2x}}

4 0
2 years ago
A cube. The top face has points G, B, C, F and the bottom face has points H, A, D, E.
Triss [41]

Answer:

its side CE

Step-by-step explanation:

I got it wrong and it showed me the answer. Hope this helps!

4 0
3 years ago
Read 2 more answers
The position of an object along a vertical line is given by s(t) = −t3 + 3t2 + 7t + 4, where s is measured in feet and t is meas
saw5 [17]

Answer:

The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

Step-by-step explanation:

Given : The position of an object along a vertical line is given by s(t) = -t^3+3t^2+7t +4, where s is measured in feet and t is measured in seconds.

To find : What is the maximum velocity of the object in the time interval [0, 4]?

Solution :

The velocity is rate of change of distance w.r.t time.

Distance in terms of t is given by,

s(t) = -t^3+3t^2+7t +4

Derivate w.r.t. time,

v(t)=s'(t) = -3t^2+6t+7

It is a quadratic function so its maximum is at vertex of the function.

The x point of the function is given by,

x=-\frac{b}{2a}

Where, a=-3, b=6 and c=7

t=-\frac{6}{2(-3)}

t=-\frac{6}{-6}

t=1

As 1 lie between interval [0,4]

Substitute t=1 in the function,

v(t)= -3(1)^2+6(1)+7

v(t)= -3+6+7

v(1)=10

Th maximum velocity is 10 ft/s.

Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.

8 0
3 years ago
Read 2 more answers
A telemarketer is successful at getting people to donate money for her organization in 55% of all calls she makes. She must get
Tems11 [23]
An interesting twist to a binomial distribution problem.

Given:
p=55%=0.55 for probability of success in solicitation
x=4=number of successful solicitations
n=number of calls to be made
P(x,n,p)>=89.9%=0.899  (from context, it is >= and not =, which is almost impossible)

From context of question, all calls are assumed independent, with constant probability of success, so binomial distribution is applicable.

The number of successes, x, is then given by
P(x)=C(n,x)p^x(1-p)^{n-x}where
p=probability of success
n=number of trials
x=number of successesC(n,x)=\frac{n!}{x!(n-x)!}

Here we need n such that
P(x,n,p)>=0.899
given
x>=4, p=0.55, which means we need to find

Method 1: if a cumulative binomial distribution table is available, we can look up n=9,10,11 and find
P(x>=4,9,0.55)=0.834
P(x>=4,10,0.55)=0.898
P(x>=4,11,0.55)=0.939
So she must make (at least) 11 calls to make sure the probability of meeting her quota is 89.9% or more.

Method 2: using technology.
Similar to method 1, we can look up the probabilities directly, for n=9,10,11
P(x>=4,9,0.55)=0.834178
P(x>=4,10,0.55)=0.8980051
P(x>=4,11,0.55)=0.9390368

Method 3: using simple calculator
Here we need to calculate the probabilities for each value of n=10,11 and sum the probabilities of FAILURE S=P(0,n,0.55)+P(1,n,0.55)+P(2,n,0.55)+P(3,n,0.55)
so that the probability of success is 1-S.
For n=10,
P(0,10,0.55)=0.000341
P(1,10,0.55)=0.004162
P(2,10,0.55)=0.022890
P(3,10,0.55)=0.074603
So that
S=0.000341+0.004162+0.022890+0.074603
=0.101995
and Probability of getting 4 successes (or more) 
=1-S
=0.898005, missing target by 0.1%

So she will have to make 11 phone calls, bring up the probability to 93.9%.  The work is similar to that of n=10.
8 0
3 years ago
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