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3 years ago

Given the function rule f(x) = x² – 4x + 3, what is the output of f(–3)? (1 point)

Mathematics

1 answer:

Vladimir [108]3 years ago

8 0

Answer:

Step-by-step explanation:

To calculate the output substitute x = - 3 into f(x), that is

f(- 3) = (- 3)² - 4(- 3) + 3 = 9 + 12 + 3 = 24 ← output

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Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean

Natali [406]

Answer:

The percentage of snails that take more than 60 hours to finish is 4.75%

The relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

There is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

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

$Z = \frac{X - \mu}{\sigma}$

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.

In this problem, we have that:

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean 50 hours and standard deviation 6 hours.

This means that $\mu = 50$ and $\sigma = 6$ .

The percentage of snails that take more than 60 hours to finish is %

The pvalue of the zscore of $X = 60$ is the percentage of snails that take LESS than 60 hours to finish. So the percentage of snails that take more than 60 hours to finish is 100% substracted by this pvalue.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{60 - 50}{6}$

$Z = 1.67$

A Zscore of 1.67 has a pvalue of .9525. This means that there is a 95.25% of the snails take less than 60 hours to finish.

The percentage of snails that take more than 60 hours to finish is 100%-95.25% = 4.75%.

The relative frequency of snails that take less than 60 hours to finish is

The relative frequence off snails that take less than 60 hours to finish is the pvalue of the zscore of X = 60.

In the item above, we find that this value is .9525.

So, the relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is:

This is the pvalue of the zscore of X = 67 subtracted by the pvalue of the zscore of X = 60. So

$X = 67$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{67 - 50}{6}$

$Z = 2.83$

A zscore of 2.83 has a pvalue of .9977.

For X = 60, we have found a Zscore o 1.67 with a pvalue of .9977

So, the percentage of snails that take between 60 and 67 hours to finish is:

$p = .9977 - 0.9525 = .0452$

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 100% subtracted by the pvalue of the Zscore of X = 76.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{76 - 50}{6}$

$Z = 4.33$

The pvalue of Z = 4.33 is 1.

So, there is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most hours.

The most hours that a snail must finish is the value of X of the Zscore when p = 0.10.

Z = -1.29 has a pvalue of 0.0985, this is the largest pvalue below 0.1. So what is the value of X when Z = -1.29?

$Z = \frac{X - \mu}{\sigma}$

$-1.29 = \frac{X - 50}{6}$

$X - 50 = -7.74$

$X = 42.26$

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical 80% of snails take between and hours to finish.

This is from a pvalue of .1 to a pvalue of .9.

When the pvalue is .1, X = 42.26.

A zscore of 1.28 is the largest with a pvalue below .9. So

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 50}{6}$

$X - 50 = 7.68$

$X = 57.68$

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

5 0

3 years ago

What are the domain and range of the function? F(x)=-3/5x^3

Ksivusya [100]

Answer:

A) Domain: All real numbers, excluding zero (-∞, 0) ∪ (0, ∞)

Range: All real numbers, excluding zero (-∞, 0) ∪ (0, ∞)

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for

4vir4ik [10]

A) $g=f'$ is continuous, so $f$ is also continuous. This means if we were to integrate $g$ , the same constant of integration would apply across its entire domain. Over $0$ , we have $g(x)=2x$ . This means that

$f_{0$

For $f$ to be continuous, we need the limit as $x\to1^-$ to match $f(1)=3$ . This means we must have

$\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2$

Now, over $x$ , we have $g(x)=-3$ , so $f_{x$ , which means $f(-5)=17$ .

b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,

$\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10$

c) $f$ is increasing when $f'=g>0$ , and concave upward when $f''=g'>0$ , i.e. when $g$ is also increasing.

We have $g>0$ over the intervals $0$ and $x>4$ . We can additionally see that $g'>0$ only on $0$ and $x>4$ .

d) Inflection points occur when $f''=g'=0$ , and at such a point, to either side the sign of the second derivative $f''=g'$ changes. We see this happening at $x=4$ , for which $g'=0$ , and to the left of $x=4$ we have $g$ decreasing, then increasing along the other side.

We also have $g'=0$ along the interval $-1$ , but even if we were to allow an entire interval as a "site of inflection", we can see that $g'>0$ to either side, so concavity would not change.

5 0

3 years ago

Help me out fast pleaseeeeeeeeeeeeeee

boyakko [2]

Answer:

81 7/12

Step-by-step explanation:

75 5/12 - (-6 1/6)= 81 7/12

8 0

3 years ago

Please help with this question!!

xxMikexx [17]

$\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2\\\\l\perp k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\\text{We have the points J(-24, -4) and K(-4, 6)}.\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{substitute:}\\\\m_1=\dfrac{6-(-4)}{-4-(-24)}=\dfrac{10}{20}=\dfrac{1}{2}\\\\\text{therefore}\ m_2=-\dfrac{1}{\frac{1}{2}}=-2\\\\\text{The formula of a midpoint:}\\\\\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)\\\\\text{substitute the coordinates of the points J and K:}$

$x=\dfrac{-24+(-4)}{2}=\dfrac{-28}{2}=-14\\\\y=\dfrac{-4+6}{2}=\dfrac{2}{2}=1\\\\\text{midpoint}\ (-14,\ 1)\\\\\text{The point-slope form:}\\\\y-y_1=m(x-x_1)\\\\\text{substitute}\ m=-2,\ x_1=-14\ \text{and}\ y_1=1:\\\\y-1=-2(x-(-14))\\\\\boxed{y-1=-2(x+14)}$

8 0

3 years ago