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3 years ago

I ate 6 pieces of my candy. This is 75% of the candy I had. How much candy did I have before

Mathematics

1 answer:

Art [367]3 years ago

6 0

Answer:

The amount of candy you had before is 8.

Step-by-step explanation:

75% + 25% makes 100% percent.

If we take 25% off from 8, you get 6.

A quick way to do this is: 8(.25)

If you solve that, you get 2. 8 - 2 is 6, which is what you originally had.

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HELP PLEASE!!! 10 PTS!!

allsm [11]

Answer:

m∠DEC = 78°

Step-by-step explanation:

Given information: AC = AD, AB⊥BD, m∠DAC = 44° and CE bisects ∠ACD.

If two sides of a triangles are congruent then the opposite angles of congruent sides are congruent.

AC = AD (Given)

$\angle ADC\cong \angle ACD$

$m\angle ADC=m\angle ACD$

According to the angle sum property, the sum of interior angles of a triangle is 180°.

$m\angle ADC+m\angle ACD+m\angle DAC=180$

$m\angle ACD+m\angle ACD+44=180$

$2m\angle ACD=180-44$

$2m\angle ACD=136$

Divide both sides by 2.

$m\angle ACD=68$

CE bisects ∠ACD.

$m\angle ACE=m\angle DCE=\dfrac{\angle ACD}{2}$

$m\angle ACE=m\angle DCE=\dfrac{68}{2}$

$m\angle ACE=m\angle DCE=34$

Use angle sum property in triangle CDE,

$m\angle CDE+m\angle DCE+m\angle DEC=180$

$68+34+m\angle DEC=180$

$102+m\angle DEC=180$

Subtract 102 from both sides.

$m\angle DEC=180-102$

$m\angle DEC=78$

Therefore, the measure of angle DEC is 78°.

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3.11 A shipment of 7 television sets contains 2 defective sets. A hotel makes a random purchase of 3 of the sets. If x is the nu

djverab [1.8K]

Answer:

Probability distribution for x:

$P(x=0)=0.3644\\\\P(x=1)=0.4373\\\\P(x=2)=0.1749\\\\P(x=3)=0.0233\\\\$

Step-by-step explanation:

We can model the number of defective sets in the group of TV sets (variable x) as a binomial variable, with sample size=3 and probability of success p=2/7≈0.2857.

The probability of k defective sets in the group is:

$P(x=k) = \dbinom{n}{k} p^{k}q^{n-k}$

So, we have this probabilty distribution for x:

$P(x=0) = \dbinom{3}{0} p^{0}q^{3}=1*1*0.3644=0.3644\\\\\\P(x=1) = \dbinom{3}{1} p^{1}q^{2}=3*0.2857*0.5102=0.4373\\\\\\P(x=2) = \dbinom{3}{2} p^{2}q^{1}=3*0.0816*0.7143=0.1749\\\\\\P(x=3) = \dbinom{3}{3} p^{3}q^{0}=1*0.0233*1=0.0233\\\\\\$

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