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Black_prince [1.1K]
3 years ago
8

Can someone help with my math question

Mathematics
1 answer:
vodka [1.7K]3 years ago
5 0
I don't see a question. )-:
You might be interested in
The price of a notebook was $3.70 yesterday. Today, the price fell to $3.20. Find the percent decrease
jeyben [28]

Answer:

86% Decrease

Step-by-step explanation:

3.70 times x= 3.20      3.20 divided by 3.70= 0.86

6 0
3 years ago
HELP++++++++++++++++++++++++
lys-0071 [83]

Answer:

ASA

ΔFGH ≅ ΔIHG ⇒ answer B

Step-by-step explanation:

* Lets revise the cases of congruence  

- SSS  ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and  

 including angle in the 2nd Δ

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ  

 ≅ 2 angles and the side whose joining them in the 2nd Δ

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles  

 and one side in the 2ndΔ  

- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse

 leg of the 2nd right angle Δ

* Lets prove the two triangles FGH and IHG are congruent by on of

 the cases above

∵ FG // HI and GH is transversal

∴ m∠FGH = m∠IHG ⇒ alternate angles

- In the two triangles FGH and IHG

∵ m∠FHG = m∠IGH ⇒ given

∵ m∠FGH = m∠IHG ⇒ proved

∵ GH = HG ⇒ common side

∴ ΔFGH ≅ ΔIHG ⇒ ASA

* ASA

 ΔFGH ≅ ΔIHG

3 0
3 years ago
Choose the best answer. *<br><br> A<br> B<br> C<br> D
viva [34]

Answer: I think its D

Explanation:I sort of made a coordinate graph

8 0
3 years ago
Read 2 more answers
Square root 15*3sqyare root 81y
Deffense [45]
<span>No simplification exists, Root remains :
</span>sqrt(15y) 
7 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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