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Liula [17]
3 years ago
15

Quadrilateral A’B’C’D’ is a dilation of quadrilateral ABCD about point P with a scale factor of 2 1/2 .

Mathematics
2 answers:
mina [271]3 years ago
8 0

Answer:

A. Enlargement

Step-by-step explanation:

We are given that,

Quadrilateral ABCD is dilated by a scale factor of 2\frac{1}{2}=\frac{5}{2}.

Now, we know that,

<h3>Dilation changes the size of the figure by the scale factor 'k'.</h3>

<em>If 'k' > 1, then the dilated figure is enlarged in size.</em>

<em>If 0 < 'k' < 1, then the dilated figure is reduced in size.</em>

Since, the given scale factor is \frac{5}{2}=2.5>1.

<h3>So, the quadrilateral ABCD is enlarged in size.</h3>

Hence, option A is correct.

kenny6666 [7]3 years ago
7 0
A it says can you tell me why??
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stiv31 [10]

Answer:

I think answer is

b. (2,1)

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5 0
3 years ago
Sherane rolls a standard, six-sided number cube.
ale4655 [162]
Answer: 1/3 or 33.3% (rounded to nearest tenth)

Step-by-step explanation:
3 and 6 are multiples of 3
6 sides of a dice
2/6= 33%
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3 years ago
What is the maximum height of the projectile?
sasho [114]

Answer:

C.226 feet

Step-by-step explanation:

On edg

5 0
3 years ago
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Two surfers and statistics students collected data on the number of days on which surfers surfed in the last month for 30 longbo
Alisiya [41]

Answer:

Do not reject H0. The mean days surfed for longboarders is significantly larger than the mean days surfed for all shortboarders

Step-by-step explanation:

The null hypothesis is that  the mean days surfed for all long boarders is larger than the mean days surfed for all short boarders

H0:  μL > μs      against the claim Ha:  μL≤ μs

the alternate hypothesis is the mean days surfed for all long boarders isless or equal to  the mean days surfed for all short boarders (because long boards can go out in many different surfing conditions)

The test statistic is

t= x1- x2/  √s1/n1+ s2/n2

1) Calculations

Longboards

Mean

ˉx=∑x/n=4+8+9+4+9+7+9+6+6+11+15+13+16+12+10+12+18+20+15+10+15+19+21+9+22+19+23+13+12+10/30

=377/30

=12.5667

Longboard Variance S2=[∑dx²-(∑dx)²/n]/n-1

=[831-(-13)²/30]/29

=831-5.6333/29

=825.3667/29

=28.4609

Shortboard Mean

ˉx=∑x/n=6+4+6+6+7+7+7+10+4+6+7+5+8+9+4+15+13+9+12+11+12+13+9+11+13+15+9+19+20+11/30

=288/30

=9.6

Shortboard Variance S2=[∑x²-(∑x)²/n]/n-1

=[ 3270-(288)2/30]/29

=3270-2764.8/29

=505.2/29

=17.4207

2) Putting values in the test statistic

t=|x1-x2|/√S²1/n1+S²2/n2

t =|12.5667-9.6|/√28.4609/30+17.4207/30

t =|2.9667|/√0.9487+0.5807

t=|2.9667|/√1.5294

t=|2.9667|/1.2367

t=2.3989

3) Degree of freedom =n1+n2-2=30+30-2=58

4) The critical region is t ≤ t(0.05) (58) =1.6716

5) Since the calculated t= 2.4 does not fall in the critical region t(0.05) (58)  ≤ 1.6716 we do not reject H0.

The p-value is 0.008969. The result is significant at p <0 .05.

6 0
3 years ago
I run out of time in 20 mins please answer I will mark you as brainliest
Maksim231197 [3]

Answer:

Hi there!

Your answer is:

A) -2

Step-by-step explanation:

f(-8) means to plug in -8 for x!

(-8)/4 =

-8/4 = -2

f(-8) = -2

7 0
3 years ago
Read 2 more answers
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