MATHEMATICS COLLEGE

80 POINTS MATHIn the diagram at the right, in which position are the tips of the scissors farther apart? Explain your reasoning..

Answers

Answer 1

Answer:

hey mate..

the position at which the tips of the scissors will be FARTHER APART is at position B.

as the angle made by the scissors at the centre is greater in position B.

Answer 2

Answer:

Answer:

the degree angle of scissor B is greater as its angle number is greater than A

32 is a greater number than 26 therefore the degree of angle 32 is wider than the degree of 26

~batmans wife dun dun dun...aka ~serenitybella

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Let be a continuous random variable that follows a normal distribution with a mean of and a standard deviation of . Find the value of so that the area under the normal curve to the right of is . Round your answer to two decimal places.

Answers

Complete Question

Let x be a continuous random variable that follows a normal distribution with a mean of 550 and a standard deviation of 75.

Find the value of x so that the area under the normal curve to the left of x is .0250.

Find the value of x so that the area under the normal curve to the right ot x is .9345.

Answer:

$x = 403$

$x = 436.75$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 550$

The standard deviation is $\sigma = 75$

Generally the value of x so that the area under the normal curve to the left of x is 0.0250 is mathematically represented as

$P( X < x) = P( (x - \mu )/( \sigma) < (x - 550 )/(75 ) ) = 0.0250$

$(X -\mu)/(\sigma ) = Z (The \ standardized \ value\ of \ X )$

$P( X < x) = P( Z < z ) = 0.0250$

Generally the critical value of 0.0250 to the left is

$z = -1.96$

=> $(x- 550 )/(75) = -1.96$

=> $x = [-1.96 * 75 ]+ 550$

=> $x = 403$

Generally the value of x so that the area under the normal curve to the right of x is 0.9345 is mathematically represented as

$P( X < x) = P( (x - \mu )/( \sigma) < (x - 550 )/(75 ) ) = 0.9345$

$(X -\mu)/(\sigma ) = Z (The \ standardized \ value\ of \ X )$

$P( X < x) = P( Z < z ) = 0.9345$

Generally the critical value of 0.9345 to the right is

$z = -1.51$

=> $(x- 550 )/(75) = -1.51$

=> $x = [-1.51 * 75 ]+ 550$

=> $x = 436.75$

Please help me with this homework equation I don’t understand and need help someone please do step by step

Answers

Step-by-step explanation:

f'(x) = 6x-12

in vertex : f'(x) = 0

6x-12 =0 => x=2

f(2) = 3 ( 2² ) -12 (2 ) +1 = -11

vertex ( 2 , -11 )

Find the LCM and solve, it's very very urgent.

Answers

Answers:

1.1050

2.1200

3.1200

4.3360

5.1080

6.480

pleaseseetheattachedpictureforfullsolution..

Hopeit helps....

Goodluck on your assignment...

A car travels 85 miles on 5 gallons of gas. Give the ratio of miles to gallons as a rate in miles per gallon.

Answers

Answer: Set up the ratio as a fraction and divide by the gallons. 90/5 = 18/1 The ratio is 18 miles/gallon.

Which of the following is a RATE?12 to 3

12 stickers to 3 students

Answers

Answer:

gfgfcvfvghcghccghcgcvchdcfgcfcfrtyc

Step-by-step explanation:

In a manual on how to have a number one song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 241.4 sec. and a standard deviation of 57.59 sec. Use a 0.05 significance level and the accompanying minitab display to test the claim that the sample is from a population of songs with a mean great thatn 210 sec. What do these results suggest about the advice given in the manual.The mini tab displays the following:

One-sample T

Test of mu=210 vs.>210

N Mean St. Dev SE Mean 95% lower bound T p

40 241.40 57.59 9.11 226.06 3.45 0.001

A H0 u>210 sec. H1 u < 210sec

B H0 u=210 sec. H1 u < 210sec

C H0 u<210 sec. H1 u> 210sec

D H0 u=210 sec. H1 u> 210sec

Identify the test statistic:

T =

Identify the P-Value

P-value=

Stat the final conclusion that addresses the original claim. Choose from below:

A. Reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec.
B. Fail to reject H0. There is insufficient evidence to support the claim that the sample is from a population of songs with a mean length great thatn 210 sec.
C. Reject H0. There is sufficient evidence to spport the claim that the sample is from a population of songs with a mean length greater than 210 sec.
D. Fail to reject H0. There is sufficient evidence to support the claim tha tthe sample is from a population of songs with a mean lenght greater than 210 sec.

What do the results suggest about the advice given in the manual?

A. The results do not suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice.
B The results suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice
C. The results suggest that 241.4 seconds is the best song lenght.
D. The results are inconclusive because the average length of a hit song is constantly changing.

Answers

Answer:

D H0 u=210 sec. H1 u> 210sec

$t=(241.4-210)/((57.59)/(√(40)))=3.448$

$p_v =P(t_((39))>3.448)=0.000684$

C. Reject H0. There is sufficient evidence to spport the claim that the sample is from a population of songs with a mean length greater than 210 sec.

Step-by-step explanation:

Data given and notation

$\bar X=241.4$ represent the sample mean

$s=57.59$ represent the sample standard deviation for the sample

$n=40$ sample size

$\mu_o =210$ 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)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is greater than 210 seconds, the system of hypothesis would be:

Null hypothesis: $\mu \leq 210$

Alternative hypothesis: $\mu > 210$

If we analyze the size for the sample is > 30 but 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:

$t=(\bar X-\mu_o)/((s)/(√(n)))$ (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".

Calculate the statistic

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

$t=(241.4-210)/((57.59)/(√(40)))=3.448$

P-value

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

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_((39))>3.448)=0.000684$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than 210 seconds.

C. Reject H0. There is sufficient evidence to spport the claim that the sample is from a population of songs with a mean length greater than 210 sec.

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