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Kamila [148]
3 years ago
8

Do you take the free samples offered in super- markets? About 70% of all customers will take free samples. Furthermore, of those

who take the free samples, about 30% will buy what they have sampled. Suppose you set up a counter in a supermarket offering free samples of a new product. The day you are offering free samples, 317 customers pass by your counter.
Mathematics
1 answer:
IgorC [24]3 years ago
8 0

Answer:

222 customers took a free sample of the product, and 67 finally bought it.

Step-by-step explanation:

Given that 317 consumers passed through the sample stand, and taking into account that 70% usually take free samples, the number of people who take free samples is determined by the following calculation:

317 x 70/100 = X

22,190 / 100 = X

221.9 = X

Therefore, of the 317 consumers, about 222 took free samples of the product. Furthermore, of those 222, it is estimated that 30% bought the product after testing the sample. To determine the number of people who bought the product, the following calculation must be made:

222 x 30/100 = X

6,660 / 100 = X

66.6 = X

Therefore, 67 consumers purchased the product after taking a free sample of it.

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masya89 [10]

Answer:

4.

l=22cm

b:16cm

h:10cm

surface area =2(lb+bh+lh)=2(22*16+16*10+22*10)

=1464cm²

5.

a=2.5dm

area cubical soda cracker box=6a²=6*2.5²=37.5dm²

6.

r=1.5dm

area of ball=4πr²=4π×1.5²=28.27dm² per ball

for 10 bal =28.27×10=282.6dm²

8 0
3 years ago
Food allergies affect an estimated 7% of children under age 5 in the US. What is the probability that in a kindergarden class of
bezimeni [28]

Answer:

Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)

Step-by-step explanation:

Given: Probability of a children under age 5 has food allergies = 7%

                                                                                                        =\frac{7}{100}

To find : Probability of atleast one of 12 student has food allergies

Probability of a chindren under age 5 does not have food allergies = 1-\frac{7}{100}

                  ⇒ prob = \frac{93}{100}

now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.

But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)

Probability of 0 student having allergy out of 12 student = (\frac{93}{100})^{12}

Therefore, Probability of atleast one of 12 student has food allergies

                  = 1-(\frac{93}{100})^{12}

                  = 0.581403702521

                  ≈ 0.58 ( approx)

                 

8 0
3 years ago
Read 2 more answers
You are the head of a building department . Your department is making final inspections of two types of new commercial buildings
stira [4]

Answer:

inspector" (and any subsequent words) was ignored because we limit queries to 32 words.

Step-by-step explanation:

hope this is correct!

5 0
3 years ago
Let U = {1, 2, 3, 4, 5, 6, 7}, A= {1, 3, 4, 6}, and B= {3, 5, 6}. Find the set A’ U B’
Art [367]

Answer:

Step-by-step explanation:

A'={2,5,7}

B'={1,2,4,7}

A'UB'={1,2,4,5,7}

8 0
3 years ago
What is [3.4(6x+2)+3]+4p-3x? PLS need help
inna [77]
Step 1: simplify

[3.4 ( 6 x + 2 ) +3] + 4 p -3 x 

Pemdas!

 6x+2=8x

3.4+3=6.4

6.4+8x=14.4

14.4x+4p=18.4xp+3= 21.4xpx 

x=8

p=18.4

Other x = 21.4
3 0
3 years ago
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