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3 years ago

Write an equation that is not a linear function

Mathematics

1 answer:

Leto [7]3 years ago

5 0

An example of a non-linear equation would be something like $x^{2} +2x+1$ which is a parabola.

Anything where there is an X with an exponent higher than 1 in a non-linear equation.

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The Appalachian Trail is a hiking trail that stretches 2158 miles from Georgia to Maine the record for completing this hike in t

inn [45]

They averaged about 17 hours a day.

3 0

3 years ago

Find the area of the figures

Mazyrski [523]

Answer:

120.

Step-by-step explanation:

area = L x W which means you multiply 15x 8.

4 0

3 years ago

Read 2 more answers

Suppose that you play the game with three different friends separately with the following results: Friend A chose scissors 100 t

Yanka [14]

Answer:

Friend A

$\hat p_A= \frac{100}{400}=0.25$

$z=\frac{0.25 -0.333}{\sqrt{\frac{0.333(1-0.333)}{400}}}\approx -3.47$

Friend B

$\hat p_B= \frac{20}{120}=0.167$

$z=\frac{0.167 -0.333}{\sqrt{\frac{0.333(1-0.333)}{120}}}\approx -3.80$

Friend C

$\hat p_C= \frac{65}{300}=0.217$

$z=\frac{0.217-0.333}{\sqrt{\frac{0.333(1-0.333)}{300}}}\approx -4.17$

So then the best solution for this case would be:

-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)

Step-by-step explanation:

Data given and notation

n represent the random sample taken

X represent the number of scissors selected for each friend

$\hat p=\frac{X}{n}$ estimated proportion of scissors selected for each friend

$p_o=\frac{1}{3}=0.333$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion that the friend will pick scissors is less than 1/3 or 0.333, the system of hypothesis would be:

Null hypothesis: $p\geq 0.333$

Alternative hypothesis: $p < 0.333$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

Friend A

$\hat p_A= \frac{100}{400}=0.25$

$z=\frac{0.25 -0.333}{\sqrt{\frac{0.333(1-0.333)}{400}}}\approx -3.47$

Friend B

$\hat p_B= \frac{20}{120}=0.167$

$z=\frac{0.167 -0.333}{\sqrt{\frac{0.333(1-0.333)}{120}}}\approx -3.80$

Friend C

$\hat p_C= \frac{65}{300}=0.217$

$z=\frac{0.217-0.333}{\sqrt{\frac{0.333(1-0.333)}{300}}}\approx -4.17$

So then the best solution for this case would be:

-3.47 (100 out of 400; 25%), -3.80 (20 out of 120; 16.7%), -4.17 (65 out of 300; 21.7%)

6 0

4 years ago

3 less than a number h is -12

soldi70 [24.7K]

H-3=-12 is the equation.
Then add 3 on both sides to solve for h.

h-3=-12
+3 +3
————-
h=-9

h-3=-12 is the equation and h=-9

7 0

4 years ago

A national dental association conducted a survey to find the average (mean) amount of time dentists spend on dental fillings per

eduard

Answer:

The probability of finding a sample mean less than 18 hours is 0.0082

Step-by-step explanation:

To find the probability of finding a sample mean less than 18 hours, we need to calculate the z-score of this sample mean 18. And the probability of finding a sample mean less than 18 hours is P(z<z(18)).

Z-score can be calculated as follows:

z(18)= $\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

X is the sample mean (18 hours)

M is the average hours dentists spend per week on fillings (20 hours)

s is the standard deviation (10 hours)

N is the sample size (144)

Putting the numbers, we get:

z(18)= $\frac{18-20}{\frac{10}{\sqrt{144} } }=-2.4$

Using z- table we can find that P(z<z(18)) = 0.0082

4 0

3 years ago