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Dvinal [7]
2 years ago
5

What are all of the x-intercepts of the continuous function in the table? (0, 8) (–4, 0) (–4, 0), (4, 0) (–4, 0), (0, 8), (4, an

swer fast plz !!!
Mathematics
1 answer:
crimeas [40]2 years ago
5 0

Answer:

The  x-intercepts of the continuous function in the table are:

(-4,0) and (4,0).

Step-by-step explanation:

We are given a table of values of the continuous functions as:

(0, 8) (–4, 0) (–4, 0), (4, 0)

The x-intercept of a function is a point on the x-axis i.e. where y=0. (i.e. The x-intercept is where a line crosses the x-axis ).

Here y=f(x).

Hence from the given set of values we have:

f(x)=0 when x= -4 and x=4.

Since we are given the points as (-4,0) and (4,0).

Hence, the x-intercepts of the function f(x) is: (-4,0) and (4,0).


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The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard
avanturin [10]

Answer:

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

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}

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that \mu = 33208, \sigma = 2503.

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when X = 33208+633 = 33841 subtracted by the pvalue of Z when X = 33208 - 633 = 32575

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57

So

X = 33841

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

Z = \frac{33841 - 33208}{357.57}

Z = 1.77

Z = 1.77 has a pvalue of 0.9616

X = 32575

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

Z = \frac{32575- 33208}{357.57}

Z = -1.77

Z = -1.77 has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

4 0
3 years ago
A structural steel fabricator cuts 25 pieces each 16.2 centimeters long from a 6.35-meter length
Daniel [21]

Answer: 2.2 meters

Step-by-step explanation:

The data is:

The initial piece is 6.35 meters long.

We cut 25 pieces of 16.2cm each, so in pieces only we have:

25*16.2cm = 405cm

1m = 100cm

Then: 405cm = (405/100) m = 4.05m

And each time that we cut a piece, we have 4mm of waste.

Then we have 25 times the 4mm of waste, this means that the total waste is:

4mm*25 = 100mm

and 1000mm = 1m

100mm = (100/1000) m = 0.1m

So the total amount that was cut is:

4.05m + 0.1m = 4.15m

The length in the channel iron will be: 6.35m - 4.15m = 2.2 meters

7 0
2 years ago
I need the answer for this exact question
denis23 [38]

Answer:

y=\frac{7}{2}x-20

Step-by-step explanation:

Let the equation of the line be y-y_1=m(x-x_1) where, 'm' is its slope and (x_1,y_1) is a point on it.

Given:

The equation of a known line is:

y=-\frac{2}{7}x+9

A point on the unknown line is:

(x_1,y_1)=(4,-6)

Both the lines are perpendicular to each other.

Now, the slope of the known line is given by the coefficient of 'x'. Therefore, the slope of the known line is m_1=-\frac{2}{7}

When two lines are perpendicular, the product of their slopes is equal to -1.

Therefore,

m\cdot m_1=-1\\m=-\frac{1}{m_1}\\m=-\frac{1}{\frac{-2}{7} }=\frac{7}{2}

Therefore, the equation of the unknown line is determined by plugging in all the given values. This gives,

y-(-6))=\frac{7}{2}(x-4)\\y+6=\frac{7}{2}x-14\\y=\frac{7}{2}x-14-6\\\\y=\frac{7}{2}x-20

The equation of a line perpendicular to the given line and passing through (4, -6) is y=\frac{7}{2}x-20.

3 0
3 years ago
PLEASE HELP ASAP!!!!!!
Paladinen [302]

its b as the slope of the line is constant so it's 2.

 

4 0
3 years ago
Read 2 more answers
Suppose you pay a dollar to roll two dice. if you roll 5 or a 6 you Get your dollar back +2 more just like it the goal will be t
LiRa [457]

Answer:

(a)$67

(b)You are expected to win 56 Times

(c)You are expected to lose 44 Times

Step-by-step explanation:

The sample space for the event of rolling two dice is presented below

(1,1), (2,1), (3,1), (4,1), (5,1), (6,1)\\(1,2), (2,2), (3,2), (4,2), (5,2), (6,2)\\(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)\\(1,4), (2,4), (3,4), (4,4), (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Total number of outcomes =36

The event of rolling a 5 or a 6 are:

(5,1), (6,1)\\ (5,2), (6,2)\\( (5,3), (6,3)\\ (5,4), (6,4)\\(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)\\(1,6), (2,6), (3,6), (4,6), (5,6), (6,6)

Number of outcomes =20

Therefore:

P(rolling a 5 or a 6)  =\dfrac{20}{36}

The probability distribution of this event is given as follows.

\left|\begin{array}{c|c|c}$Amount Won(x)&-\$1&\$2\\&\\P(x)&\dfrac{16}{36}&\dfrac{20}{36}\end{array}\right|

First, we determine the expected Value of this event.

Expected Value

=(-\$1\times \frac{16}{36})+ (\$2\times \frac{20}{36})\\=\$0.67

Therefore, if the game is played 100 times,

Expected Profit =$0.67 X 100 =$67

If you play the game 100 times, you can expect to win $67.

(b)

Probability of Winning  =\dfrac{20}{36}

If the game is played 100 times

Number of times expected to win

=\dfrac{20}{36} \times 100\\=56$ times

Therefore, number of times expected to loose

= 100-56

=44 times

8 0
3 years ago
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