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2 years ago

Which descriptions from the list below accurately describe the relationship between XYZ and UVW? Check all that apply.

Mathematics

2 answers:

krek1111 [17]2 years ago

5 0

Answer:

The answers are both A. Same Size and C. Similar.

Step-by-step explanation:

We know this because they are firstly both triangles. These are the same shape portion.

We also know they are similar because they have all of the same angles, just not the same side lengths.

Guest

1 year ago

YERP

Kisachek [45]2 years ago

5 0

Answer:

The correct options are A and C.

Step-by-step explanation:

In triangle XYZ and UVW,

$\angle Y=\angle V=37^{\circ}$

$\angle Z=\angle W=53^{\circ}$

By AA property of similarity, if two corresponding angles of triangle are same then both triangles are similar.

$\triangle XYZ=\triangle UVW$

The triangle XYZ and UVW have same shape and these are similar.

Therefore, the correct options are A and C.

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Rasheed wants to buy two new video games that cost $28.50 each. He earns money for them by mowing lawns in his neighborhood, and

myrzilka [38]

Answer:

$15.25

Step-by-step explanation:

$7.50+$4.85+$12.25 +$9.00 +$8.15 = $41.75

$28.50*2 = $57.00

So, he needs $57.00- $41.75 = $15.25 more

4 0

2 years ago

Read 2 more answers

If a 100-lb. cake of ice is placed on a plane of which the inclination to the horizontal is 35, what force will be required to k

Sliva [168]

F = 100(.5736)
= 57.36 lbs. (rounded off to 2 decimal places) 

2) sin60 = .866 
F = 18(.866) 
= 15.59 lbs. (rounded off to 2 decimal places)

8 0

2 years ago

An experiment was performed to compare the abrasive wear of two different laminated materials. Twelve pieces of material 1 were

vladimir1956 [14]

Answer:

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units

Step-by-step explanation:

Step:-(1)

Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4

Mean of the first sample x₁⁻ =85

standard deviation of the first sample S₁ = 4

Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.

Mean of the first sample x₂⁻ =81

standard deviation of the first sample S₂ = 5

Step :-2

Null hypothesis: H₀: there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Alternative hypothesis :H₁: there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²

The test statistic

$Z= \frac{x_{1} -x_{2} }{\sqrt{\frac{S^2_{1} }{n_{1} } +\frac{S^2_{2} }{n_{2} } } }$

Given n₁=n₂=60.

$Z= \frac{85-81 }{\sqrt{\frac{(4)^2 }{60 } +\frac{5^2 }{60} } }$

On calculation, we get

Z = $\frac{4}{\sqrt{0.6833} }$

z = 4.8389

The tabulated value Z =1.96 at 0.05 level of significance.

The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.

The null hypothesis is rejected.

Conclusion:-

there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.

3 0

2 years ago

77 million to 200.2 million

Luda [366]

77 million to 200.2 millionFrom 77 million it becomes 200.2 million.It is very noticeable that it increased we only need to identify the number that has increased.=> 200.2 - 77 million = 123.2 millionNow, let's find the increase rate:=> 123.2 million / 200.2 million = 0.62Now let's convert this to percentage=> 0.62 * 100% = 62%

4 0

2 years ago

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a)

UNO [17]

Answer:

a) P=0.226

b) P=0.6

c) P=0.0008

d) P=0.74

Step-by-step explanation:

We know that the seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Therefore, we have 46 balls.

a) We calculate the probability that are 3 red, 2 blue, and 2 green balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_3^{12}\cdot C_2^{16}\cdot C_2^{18}=660\cdot 120\cdot 153=12117600$

Therefore, the probability is

$P=\frac{12117600}{53524680}\\\\P=0.226$

b) We calculate the probability that are at least 2 red balls.

We calculate the probability withdrawn of 1 or none of the red balls.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations: for 1 red balls

$C_1^{12}\cdot C_7^{34}=12\cdot 1344904=16138848$

Therefore, the probability is

$P_1=\frac{16138848}{53524680}\\\\P_1=0.3$

We calculate the number of favorable combinations: for none red balls

$C_7^{34}=5379616$

Therefore, the probability is

$P_0=\frac{5379616}{53524680}\\\\P_0=0.1$

Therefore, the the probability that are at least 2 red balls is

$P=1-P_1-P_0\\\\P=1-0.3-0.1\\\\P=0.6$

c) We calculate the probability that are all withdrawn balls are the same color.

We calculate the number of possible combinations:

$C_7^{46}=\frac{46!}{7!(46-7)!}=53524680$

We calculate the number of favorable combinations:

$C_7^{12}+C_7^{16}+C_7^{18}=792+11440+31824=44056$

Therefore, the probability is

$P=\frac{44056}{53524680}\\\\P=0.0008$

d) We calculate the probability that are either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Let X, event that exactly 3 red balls selected.

$P(X)=\frac{C_3^{12}\cdot C_4^{34}}{53524680}=0.57\\$

Let Y, event that exactly 3 blue balls selected.

$P(Y)=\frac{C_3^{16}\cdot C_4^{30}}{53524680}=0.29\\$

We have

$P(X\cap Y)=\frac{18\cdot C_3^{12} C_3^{16}}{53524680}=0.12$

Therefore, we get

$P(X\cup Y)=P(X)+P(Y)-P(X\cap Y)\\\\P(X\cup Y)=0.57+0.29-0.12\\\\P(X\cup Y)=0.74$

8 0

2 years ago