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notka56 [123]
4 years ago
6

In a class 70% passed in mathematics and 80 % passed in English. Together 40% failed in maths and English in a class of 133 memb

ers. How many students passed in both the subjects
Mathematics
1 answer:
geniusboy [140]4 years ago
4 0
For this specific problem, this would be the solution:   <span><span>= 0.7(133) + 0.8(133) - 2(0.4[133]) </span><span>
<span>= 93 + 106 - 2(53) </span>
<span>= 199 - 106 </span>
<span>= 93 </span></span></span>

This means 93 students passed in both the subjects. I am hoping that this answer has satisfied your query about and it will be able to help you.

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Given the following diagram, enter the required information.
viva [34]

Answer:

  • 39° 44' 52''

Step-by-step explanation:

<h3>Given</h3>
  • m∠DXB = 70° 15' 12''
  • m∠DXC = 30° 30' 20''
<h3>To find</h3>
  • m∠CXB
<h3>Solution</h3>

<u>According to Angle Addition postulate:</u>

  • m∠DXB = m∠DXC + m∠CXB

<u>Therefore</u>

  • m∠CXB = m∠DXB - m∠DXC
  • m∠CXB = 70° 15' 12''- 30° 30' 20'' = 39° 44' 52''
7 0
3 years ago
A bakery received a shipment of 6,562 peaches. If it takes 6 peaches to bake a peach pie, about how many peach pies can the bake
chubhunter [2.5K]
1094

6562/6 = 1093.67 ~ 1094


Mark brainliest please
4 0
3 years ago
When an expression is in its simplest form it has no _________ or _______ ________
CaHeK987 [17]
I don’t know the answer I just need some points have a good day
6 0
3 years ago
2.63 According to Consumer Digest (July/August 1996), the probable location of personal computers (PC) in the home is as follows
PolarNik [594]

Answer:

a) 0.32

b) 0.68

c) office or den

Step-by-step explanation:

Locations             Probabilities

Adult bedroom    0.03

Child bedroom    0.15

Other bedroom   0.14

Office or den       0.40

Other rooms        0.28

a)

P(PC in bedroom)= P(PC in adult bedroom)+ P(PC in child bedroom)+ P(PC in other bedroom)

P(PC in bedroom)= 0.03+0.15+ 0.14

P(PC in bedroom)= 0.32.

Thus, the probability that a PC is in a bedroom is 0.32.

b)

P(PC is not in bedroom)= P(PC in Office or den)+ P(PC in Other rooms)

P(PC is not in bedroom)= 0.40+0.28

P(PC is not in bedroom)= 0.68.

Thus, the probability that a PC is not in a bedroom is 0.68.

c)

When a household is selected at random from households with a PC we would expect to find a PC in a room which has a greater probability of having PC.

The greater probability of room having a PC is of Office or den room with probability 0.40. So, when a household is selected at random from households with a PC we would expect to find a PC in a Office or den room.

3 0
3 years ago
How many distinct products can be formed using two different integers from the given set: {–6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4
zhannawk [14.2K]

Number of distinct products that can be formed is 144

<h3>Permutation</h3>

Since we need to multiply two different integers to be selected from the set which contains a total of 12 integers. This is a permutation problem since we require distinct integers.

Now, for the first integer to be selected for the product, since we have 12 integers, it is to be arranged in 1 way. So, the permutation is ¹²P₁ = 12

For the second integer, we also have 12 integers to choose from to be arranged in 1 way. So, the permutation is ¹²P₁  = 12.

<h3>Number of distinct products</h3>

So, the number of distinct products that can be formed from these two integers are ¹²P₁ × ¹²P₁ = 12 × 12 = 144

So, the number of distinct products that can be formed is 144

Learn more about permutation here:

brainly.com/question/25925367

6 0
3 years ago
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