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3 years ago

8

Describe the process you would use to create the perpendicular bisector to a segment XY using only an unmarked straightedge and

an unmarked compass pls help

1 answer:

laiz [17]3 years ago

3 0

Please reference the attached diagram.

Set the compass to a radius greater than half the distance between X and Y
Mark arcs on either side of the segment XY using X as a center
Mark arcs on either side of the segment XY using Y as a center
Draw a line (JK) between the points where the arcs intersect. This is your perpendicular bisector.

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If a 24-kg mass stretches a spring 15 cm, what mass will stretch the spring 10 cm

natulia [17]

Answer:

<h2>16kg</h2>

Step-by-step explanation:

This problem is borers on elasticity of materials.

according to Hooke's law,<em> "provided the elastic limit of an elastic material is not exceeded the the extension e is directly proportional to the applied force."</em>

$F= ke$

where F is the applied force in N

k is the spring constant N/m

e is the extension in meters

Given data

mass m= 24kg

extensnion=15cm in meters= $\frac{15}{100}$ = $0.15m$

we can solve for the spring constant k

we also know that the force F = mg

assuming $g=9.81m/s^{2}$

therefore

$24*9.81=k*0.15\\235.44=k*0.15\\k=\frac{235.44}{0.15} \\k=1569.6N/m\\$

We can use this value of k to solve for the mass that will cause an extension of $10cm= 0.1m$

$x*9.81=1569.6*0.1\\\\x= \frac{156.96}{9.81} \\\x= 16kg$

5 0

3 years ago

1. Billy's mother started a college fund on Billy's 8th birthday. She

MaRussiya [10]

When billy turns 18 he will have $3,099.6 in his account because .021 percent x $8,200= $172.20 per year x 18 years = $3,099.60

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2 years ago

There are 16 players in my poker group that meets the first Tuesday of every month. Ten of the players are male and six are fema

rodikova [14]

Answer: Hello mate!

in your group, there are 16 players, 10 males, and 6 females.

in this case, 5 males already arrived (then there are 5 other males left)

and 4 females already arrived (then there are other 2 females left)

then there are a total of 7 players left, where 2 are female and 5 are males.

you want to know the probability that the next person through the door will be a male.

this is the number of male players left, divided by the number of players left: 5/7 = 0.71

then the probability that the next person through the door will be a male is 0.71.

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3 years ago

The base of a pyramid has n sides.<br> Write an expression for the number of faces of the pyramid.

Naya [18.7K]

N+1
Ex: A square pyramid has 5 faces, the base of the square pyramid has 4 sides.
4+1=5

7 0

3 years ago

Consider the following data set. The data were actually collected in pairs, and each row represents a pair.

Anna11 [10]

Answer:

Step-by-step explanation:

Hello!

a.

Using the data sets you have to analyze them as if they are two independent samples.

The hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

Assuming that both data sets are from a normal distribution and both population variances, although unknown, are equal, the statistic to use is:

$t=\frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}$

$Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}$

$Sa^2= \frac{9*5.38+9*3.65}{10+10-2} =4.51$

Sa= 2.12

$t_{H_0}= \frac{(49.69-50.52)-0}{2.12*\sqrt{\frac{1}{10} +\frac{1}{10} } } = -0.875$

The p-value for the two-tailed test is:

P( $t_{18}$ ≤-0.875) + P( $t_{18}$ ≥0.875)= P( $t_{18}$ ≤-0.875) + (1 - P( $t_{18}$ ≤0.875))= 0.1965 + ( 1 - 0.8035)= 0.393

Since there is no significant level I determined it at 5%, comparing it to the p-value, the hypothesis test is not significant. Meaning that the difference between the population means of both groups is equal to zero.

b.

Considering the given data as a paired sample, you have to determine the variable difference to conduct the test:

Xd: the difference between X₁ and X₂

Before the sample mean and sample standard deviation you have to calculate the difference between the observations of group 1 and group 2. (2nd attachment)

Mean X[bar]d= -0.83

Standard deviation Sd= 1.28

The hypotheses for the paired sample test are:

H₀: μd=0

H₁: μd≠0

α: 0.05

$t= \frac{X[bar]d-Mud}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}$

$t_{H_0}= \frac{-0.83-0}{\frac{1.28}{ \sqrt{10} }} = -2.05$

The p-value for the two-tailed test is:

P( $t_{9}$ ≤-2.05) + P( $t_{9}$ ≥2.05)= P( $t_{9}$ ≤-2.05) + (1 - P( $t_{9}$ ≤2.05))= 0.0353 + (1 - 0.9647)= 0.0706

Comparing the p-value with the significance level, the hypothesis test is not significant. Meaning that the population mean of the difference between group 1 and group 2 is equal to cero. There is no difference between the two groups.

c.

The value of the statistic in "a" is greater than the value of the statistic obtained in "b". Since there are two samples used an "a" the degrees of freedom of the test are the double as the ones used in "b". The p-value obtained in "b" is around half of the p-value of "a".

The main difference is that in "a" you compared two different samples but in "b" you compared two dependent samples when analyzing paired data the "effect of the individual" is removed from the equation.

I hope this helps!

3 0

3 years ago