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yulyashka [42]
3 years ago
6

Find the zero of f(x)=-3x-18

Mathematics
1 answer:
navik [9.2K]3 years ago
5 0

Answer:

2 and -3

Step-by-step explanation:

The reason why these are the answers is because if you put it in your graphics calculator then you can see the only points where the diagram intersects with the x-axis would be at points 2 and -3 :)

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3 inches equals how many yards?
Stels [109]

Answer:\

3 inches does not even equal a whole yard it equals 0.08 of a yard

6 0
3 years ago
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Paolo says that his bicycle is hard to pedal. Mia looks at the bicycle and tells him he needs to oil the chain. Explain why oili
fgiga [73]
It would help the chain run smoother and work better and faster.
6 0
3 years ago
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a survey amony freshman at a certain university revealed that the number of hours spent studying the week before final exams was
Marat540 [252]

Answer:

Probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

Step-by-step explanation:

We are given that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15.

A sample of 36 students was selected.

<em>Let </em>\bar X<em> = sample average time spent studying</em>

The z-score probability distribution for sample mean is given by;

          Z = \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} }  ~ N(0,1)

where, \mu = population mean hours spent studying = 25 hours

            \sigma = standard deviation = 15 hours

            n = sample of students = 36

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the average time spent studying for the sample was between 29 and 30 hours studying is given by = P(29 hours < \bar X < 30 hours)

    P(29 hours < \bar X < 30 hours) = P(\bar X < 30 hours) - P(\bar X \leq 29 hours)

      

    P(\bar X < 30 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } < \frac{ 30-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z < 2) = 0.97725

    P(\bar X \leq 29 hours) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} } }} } \leq \frac{ 29-25}{\frac{15}{\sqrt{36} } }} } ) = P(Z \leq 1.60) = 0.94520

                                                                    

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 and x = 1.60 in the z table which has an area of 0.97725 and 0.94520 respectively.</em>

Therefore, P(29 hours < \bar X < 30 hours) = 0.97725 - 0.94520 = 0.0321

Hence, the probability that the average time spent studying for the sample was between 29 and 30 hours studying is 0.0321.

7 0
3 years ago
Natalia sawed five boards of equal links to make a school each was 9/10 of a meter long what is the total length of the board sh
Alexxandr [17]
5×9/10=4.5
        9/10          1 8/10         2 7/10        3 6/10             4 5/10
---------------l--------------l---------------l----------------l----------------
5 5/10=4 1/2=4.5
6 0
3 years ago
Evaluate the difference quotient for the given function. Simplify your answer.
nasty-shy [4]

I suppose you mean

f(x) = \dfrac{x+5}{x+1}

Then

f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2

and the difference quotient is

\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}

If it's the case that <em>x</em> ≠ 3, then (<em>x</em> - 3)/(<em>x</em> - 3) reduces to 1, and you would be left with

\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}

4 0
2 years ago
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