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2 years ago

Please let me know if you guys can get this.!

Mathematics

1 answer:

geniusboy [140]2 years ago

4 0

Answer:

Step-by-step explanation:

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It has been reported that men are more likely than women to participate in online auctions. In a recent survey, 65% of repondent

kupik [55]

Answer:

Hence, the Female and not participated is $\frac{28}{100}=0.28\%$

Step-by-step explanation:

Let Total respondents $=100$

Participated in online action $(P)=65$

Men $(m)=45$ and Women $(F)=100-45=55$

Male and participated $(M\; \text{and}\;P)=38$

Female and participated $(F\; \text{and}\;P)=65-38=27$

Male and not participated is $45-38=7$

Female and not participated is $55-27=28$

Therefore, the Female and not participated is $\frac{28}{100}=0.28\%$

Hence, the correct option is $D$ .

5 0

2 years ago

AC=5x-4 and CD=16-3x solve for x

mash [69]

X=-13
;;^^^^^^%:^%%%%%%%

3 0

3 years ago

If x+9.8=14.7, what is the value of 8(x-3.7?

yulyashka [42]

First find X by subtracting 9.8 from 14.7
X should equal 4.9
than fill in the equation for x
8(4.9-3.7)
you then subtract in the parenthasis
getting the answer of 1.2
finally multiply 1.2x8
Your final answer is 9.6

3 0

2 years ago

A surfboard has an original price of $259. It is on sale for 55% off the original price.

grin007 [14]

I'm not sure what answer you need for this question?
original price: 259
sale: 55% off
55%=0.55
so 259×0.55=142.45
142.45 is the 55% of the original price
259－142.45=116.55
116.55 is the price after sale.
hope this would help. (*^ω^*)/

5 0

2 years ago

Read 2 more answers

There are eight students in a class. Only one of them has passed Exam P/1 and only one of them has passed Exam FM/2. No student

Svetach [21]

Answer: There is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

Step-by-step explanation:

Total number of students = 8

Number of student who has passed Exam P/1 = 1

Number of student who has passed Exam FM/2 = 1

No student has passed more than one exam.

According to question, exactly three students from a randomly chose group of four students have not passed Exam P/1 or Exam FM/2.

So, Probability will be

$\frac{^6C_3\times ^2C_1}{^8C_4}\\\\=\frac{20\times 2}{70}\\\\=\frac{4}{7}\\\\=0.57$

Hence, there is probability of 0.57 chances that exactly three students from a group of four students have not passed Exam P/1 or Exam FM/2.

4 0

3 years ago