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Gelneren [198K]
3 years ago
7

If a student answer 67 exam questions correctly out of 100,what fraction and what personage of question did the students answers

incorrectly.
A. 67/100
B.67%
C. 0.67
D. 30/100
E. 33 %
F. 33/100
Mathematics
2 answers:
katovenus [111]3 years ago
6 0
I think it’s A and B
VikaD [51]3 years ago
5 0

Answer:

A and D

Step-by-step explanation:

67 right out of 100 for the fraction and 67/100 = 67%

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Please hurry. it's geometry.
Ainat [17]

That would be option A as the angles  and sides  ( the AS) have already been stated.

7 0
3 years ago
One bar of candy A and two bars of candy B have 774 calories. Two bars of candy A and one bar of candy B contains 786 calories.
vampirchik [111]
◆ Define the variables:

Let the calorie content of Candy A = a
and the calorie content of Candy B = b

◆ Form the equations:

One bar of candy A and two bars of candy B have 774 calories. Thus:

a + 2b = 774

Two bars of candy A and one bar of candy B contains 786 calories

2a + b = 786

◆ Solve the equations:

From first equation,
a + 2b = 774
=> a = 774 - 2b

Put a in second equation
2×(774-2b) + b = 786
=> 2×774 - 2×2b + b = 786
=> 1548 - 4b + b = 786
=> -3b = 786 - 1548
=> -3b = -762
=> b = -762/(-3) = 254 calorie

◆ Find caloric content:

Caloric content of candy B = 254 calorie

Caloric content of candy A = a = 774 - 2b = 774 - 2×254 = 774 - 508 = 266 calorie
5 0
3 years ago
Which of the following fractions is not equivalent to 6/21
andreev551 [17]

Answer:

3/7 and 4/14

Step-by-step explanation:

a) 3/7 is not an equivalent of 6/21 because when you multiply the numerator and the denominator by 2 you get: 6/14

b) 4/14 is not an equivalent

4 0
3 years ago
Read 2 more answers
3.
Naddika [18.5K]
It would take 630 seconds to boil the water.

7 0
3 years ago
In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH
solniwko [45]

Answer:

See explanation

Step-by-step explanation:

In ΔABC, m∠B = m∠C.

BH is angle B bisector, then by definition of angle bisector

∠CBH ≅ ∠HBK

m∠CBH = m∠HBK = 1/2m∠B

CK is angle C bisector, then by definition of angle bisector

∠BCK ≅ ∠KCH

m∠BCK = m∠KCH = 1/2m∠C

Since m∠B = m∠C, then

m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH   (*)

Consider triangles CBH and BCK. In these triangles,

  • ∠CBH ≅ ∠BCK (from equality (*));
  • ∠HCB ≅ ∠KBC, because m∠B = m∠C;
  • BC ≅CB by reflexive property.

So, triangles CBH and BCK are congruent by ASA postulate.

Congruent triangles have congruent corresponding sides, hence

BH ≅ CK.

5 0
3 years ago
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