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3 years ago

Estimate the values of x when y=1

Mathematics

1 answer:

Cerrena [4.2K]3 years ago

7 0

Step-by-step explanation:

if one were to put the y value the equation they could find out what x is for a certain value of y

putting the value of y(1), we get the equation

1=4x-1

2=4x

x=0.5

the value of x when y is 1 is =0.5

NOTE: I got this explanation on brainy on someone else I am not taking the credit

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What is the percent of change if 90 is decreased to 82

Sedbober [7]

Answer: - 8.89%

I am going to explain you in detail ir order to help you to be able to do this kind of problems by yourself.

The definition of percent change, %, is:

          change
% = ---------------- * 100
        base value

The base value is from where you start or the value for which you want to asses the variation.

So, in this case the base value is 90 (starting value).

The change is final value - initial value = 82 - 90 = - 8. The negative sign means that the value decreased.

So, following the formula you get:

         82 - 90                - 8
% = ----------- * 100 = -------- * 100 = - 8.89
            90                     90

Answer: the percent of change is - 8.89 %, meaning that the value decreased 8.89%.

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3 years ago

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Please help!! MATH | WILL MARK BRAINLIEST!! 40 POINTS

madreJ [45]

Answer:

y = 9 cos [(⅕pi)x]

Step-by-step explanation:

Amplitude = 9

Period = 10

2pi/10 = pi/5

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2 years ago

What’s the point of the tangency

lbvjy [14]

M is your answer, because the tangent intersects the circle at point M.

So M is correct.

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3 years ago

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The total cost of renting a banquet hall is a function of the number of hours the hall is rented. The owner of the banquet hall

madam [21]

Answer:

It would be $390 for the 4 hours and the cleaning fees. And the slope would be -13 if you are going from 12 to 7 but would be 13 if you are going from 7 to 12...

Step-by-step explanation:

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2 years ago

Assume that females have pulse rates that are normally distributed with a mean of μ=73.0 beats per minute and a standard deviati

Gennadij [26K]

Answer:

a. the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, the probability that they have pulse rates with a mean less than 76 beats per minute is 0.8849

c. D. Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

Step-by-step explanation:

Given that:

Mean μ =73.0

Standard deviation σ =12.5

a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 76 beats per minute.

Let X represent the random variable that is normally distributed with a mean of 73.0 beats per minute and a standard deviation of 12.5 beats per minute.

Then : X $\sim$ N ( μ = 73.0 , σ = 12.5)

The probability that her pulse rate is less than 76 beats per minute can be computed as:

$P(X < 76) = P(\dfrac{X-\mu}{\sigma}< \dfrac{X-\mu}{\sigma})$

$P(X < 76) = P(\dfrac{76-\mu}{\sigma}< \dfrac{76-73}{12.5})$

$P(X < 76) = P(Z< \dfrac{3}{12.5})$

$P(X < 76) = P(Z< 0.24)$

From the standard normal distribution tables,

$P(X < 76) = 0.5948$

Therefore , the probability that her pulse rate is less than 76 beats per minute is 0.5948

b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 76 beats per minute.

now; we have a sample size n = 25

The probability can now be calculated as follows:

$P(\overline X < 76) = P(\dfrac{\overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{ \overline X-\mu}{\dfrac{\sigma}{\sqrt{n}}})$

$P( \overline X < 76) = P(\dfrac{76-\mu}{\dfrac{\sigma}{\sqrt{n}}}< \dfrac{76-73}{\dfrac{12.5}{\sqrt{25}}})$

$P( \overline X < 76) = P(Z< \dfrac{3}{\dfrac{12.5}{5}})$

$P( \overline X < 76) = P(Z< 1.2)$

From the standard normal distribution tables,

$P(\overline X < 76) = 0.8849$

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

In order to determine the probability in part (b); the normal distribution is perfect to be used here even when the sample size does not exceed 30.

Therefore option D is correct.

Since the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

5 0

2 years ago