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2 years ago

How do you write an equation from a line of best fit on a graph?

Mathematics

1 answer:

marshall27 [118]2 years ago

8 0

Answer:

find y intercept and slope

Step-by-step explanation:

You might be interested in

According to a survey, 65% of murders committed last year were cleared by arrest or exceptional means. Fifty murders committed

monitta

Answer:

a) $P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421$

b) $P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714$

$P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502$

$P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319$

And adding these values we got:

$P(36 \leq X \leq 38)= 0.1535$

c) We can find the expected value given by:

$E(X) = np =50*0.65 = 32.5$

And the standard deviation would be:

$\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373$

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

$\mu -2\sigma = 32.5- 2*3.373 = 25.75$

And then we can consider a value of 18 as unusual lower for this case.

Step-by-step explanation:

Let X the random variable of interest "number cleared by arrest or exceptional", on this case we can model this variable with this distribution:

$X \sim Binom(n=50, p=0.65)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421$

Part b

We want this probability:

$P(36 \leq X \leq 38)$

We can find the individual probabilities:

$P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714$

$P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502$

$P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319$

And adding these values we got:

$P(36 \leq X \leq 38)= 0.1535$

Part c

We can find the expected value given by:

$E(X) = np =50*0.65 = 32.5$

And the standard deviation would be:

$\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373$

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

$\mu -2\sigma = 32.5- 2*3.373 = 25.75$

And then we can consider a value of 18 as unusual lower for this case.

6 0

3 years ago

Determine whether f(x) = 5x+1/x and g(x)= x/ 5x+1 are inverse functions explain how you know. please explain in steps or worded

Len [333]

Hello from MrBillDoesMath!

Answer:

Discussion:

If the functions are inverses, then f(g(x)) = x for appropriate x. Consider x = 1.

g(1) = 1/ (5(1) + 1) = 1/6

f(g(1)) = f (1/6) = 5(1/6) + 1/ (1/6) = 5/6 + 6 = 6 5/6 <> 1

Conclusion: f(g(1)) <> 1 so the functions are not inverses

Thank you,

MrB

7 0

3 years ago

The length of two sides of a triangle are 10, 15 find the range of possible lengths for the third side

Zepler [3.9K]

Answer:

Range for third side is

( 5 , 25 ) cm.

Step-by-step explanation:

As two sides of triangle are 10 and 15 ,

the third side would have to be less than the sum of other two sides i.e. less than 25 cm.

On the other hand if it is smaller one than this side plus side of length 10

should be greater than 15 and therefore

this side is greater than 15 − 10 = 5 cm.

Hence range is

( 5 , 25 )

Hope this answer helps you :)

Have a great day

Mark brainliest

5 0

2 years ago

X-3= 0 and 4x+2y-1= 0

Assoli18 [71]

X-3=0
x=3

4x+2y-1=0
4(3)+2y-1=0
12+2y-1=0
11+2y=0
2y=-11
y=-5.5

3 0

3 years ago

Read 2 more answers

The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o

timurjin [86]

Answer:

a) Null hypothesis: $\mu = 90000$

Alternative hypothesis: $\mu \neq 90000$

b) $p_v =2*P(t_{(24)}$

c) If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

Step-by-step explanation:

1) Data given and notation

77600 ,76000 ,90700 ,97200 ,90700 ,101800 ,78700 ,81300 ,84200 ,97600 ,

77500 ,75700 ,89400 ,84300 ,78700 ,84600 ,87700 ,103400 ,83800 ,101300

94700 ,69200 ,95400 ,61500 ,68800

We can calculate the sample mean and deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

The values obtained are:

$\bar X=85272$ represent the mean annual salary for the sample

$s=11039.23$ represent the sample standard deviation for the sample

$n=25$ sample size

$\mu_o =90000$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean salary differs from 90000, the system of hypothesis would be:

Null hypothesis: $\mu = 90000$

Alternative hypothesis: $\mu \neq 90000$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b: Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{85272-90000}{\frac{11039.23}{\sqrt{25}}}=-2.141$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(24)}$

Part c: Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean for the salary differs from 9000 at 5% of significance.

6 0

3 years ago